4. Integration. Type - I Integration of a proper algebraic rational function r(x) = p(x), with nonrepeated real linear factors in the denominator.

Transcription

1 4. Integration The process of determining an integral of a function is called integration and the function to be integrated is called integrand. The stu of integral calculus consists in developing techniques for the determination of integral of a given function. This finds etensive applications to geometry, nature and social sciences. In this chapter we shall be laying emphasis mainly on the different methods of integration. The process of integration is largely of a tentative nature and is not so systematic as that of differentiation. In general, eperience is the best guide for suggesting the quickest and the simplest method for integrating a given function. Integration of rational functions by using partial fractions: In this section we shall be concerned with the integration of rational functions f() where f() and g() are polynomials. We use the fact that every g() real polynomial can be factored into real factors of the first and second degree. Type - I Integration of a proper algebraic rational function r() = p(), with nonrepeated real linear factors in the denominator. q() Step - I: Let q() = a + bc + d l + m. From the equation p() q() = A + B + obtain a polynomial identity by multiplying both sides a+b c+d by q(). Step - II: Comparing coefficients of powers of gives a system of linear equations. Step - III: Solve this system of equations to obtain constants A, B, C, Step - IV: Substitute all the values of A, B, back in the equation p() A + a+b b + c. B + and integrate by using the formula c+d Eample : Evaluate 9. a+b = a q() = log a +

2 Solution: = = 6 9 ( )(+) 9 = 6 6 ). + + ( 6 + = 6 ln ln + + c 6 = (ln ln + ) + c 6 = 6 ln + + c In general = ln a a a +a + c. Eample : Evaluate. (+)(+) Solution: Let = A + B. (+)(+) (+) + = A( + ) + B( + ) Put =. Put =. Thus = A( + ) + B( + ) = A() + B(0) = A. A =. = A( + ) + B( + ) = A(0) + B( ) = B B =. = +. (+)(+) (+) + ( + )( + ) = = ln + ln + + c = ln c.

3 Eample : Evaluate + (+6)( 5). + Solution: Let + = A( 5) + B( + 6). Put = 6. ( 6) + = A( 6 5) + B(0) A = + = 9 (+6)( 5) = A = 9. Put = 5. (5) + = A(0) + B(5 + 6). B = B =. + = 9 (+6)( 5) +6 = 9 ln ( + 6) + A + B ln 5 + c. 5 Eample 4: Evaluate ( )(+)( ). Solution: Let ( )(+)( ) + = A( + )( ) + B( )( ) + C( )( + ) Put =. = A( + ) ( ) + B(0) + C(0) 4A = A =. 4 Put =. ( ) = A(0) + B( ) ( ) + C(0) ( )( 4)B = 8B = B =. 8 Put =. = A(0) + B(0) + C( )( + ) ()(4)C = 9 8C = 9 C = 9. 8 ( )(+)( ) = 4 = 4 ln + 8 ln ln Integration of an improper rational function with non-repeated linear factors in the denominator. Step - : Apply division algorithm to obtain p() = q() Quotient + remainder.

4 4 Step - : Consider p() q() = Quotient + remainder q(). Step - : p() = Quotient + remainder. q() q() Step - 4: Integrate the rational function, remainder. q() by previous method, as it is a proper rational function. Eample : Evaluate 7 +6 (+)( )( ). Solution: Here ( + )( )( ) = ( + )( )( ) ( + )( )( ) = +4 = (+)( )( ) = = + ( + )( )( ) ( + )( )( ) + 4 = + () ( + )( )( ) Let A + B + C = A( )( ) + B( + )( ) + C( + )( ). Put = ( ) + 4( ) = A( )( ) + B(0) + C(0) 4 = A( )( 4) 6 = A A = 6 =. Similarly putting =,, we get B = 7, C =. +4 = + 7 (+)( )( ) + = ln ln ln, by equation (), 7 +6 = ln ln (+)( )( ) ln.

5 5 Integration of an algebraic rational function with repeated linear factors in the denominator Step - : For the repeated linear factor of the form (a+b) n in the denominator, assign corresponding partial fractions as A + A + + An. a+b (a+b) (a+b) n Step - : Assign partial fractions to the non-repeated linear factors thus B + B +. c+d c+f Step - : Consider the equation r() = p() = A + A L + + An + B q() a+b (a+b) (a+b) n + B + c+d c+f From the above equation obtain a polynomial identity by multiplying both sides by q(). Step - 4: Compare coefficients of powers of and solve for A, A,, A n, B, B,. Step - 5: Put back the values of A, A,, A n, B, B, in step, and integrate. Use = ln a + b + c, a+b a For k, = (a + b) k = (a+b) k+ =. (a+b) k ( k+)a ( k)a(a+b) k Eample : Evaluate 4 (+). Solution: Let 4 (+) = A + + B (+) + C 4 = A( + ) + B + C( + ) 4 = A + A + B + C( ) = A + A + B + C + 4C + 4C = (A + C) + (A + B + 4C) + 4C Comparing coefficients, A + C =, A + B + 4C =, 4C = 4, C = A =. + B + 4( ) = 4 + B 4 =.

6 6 B =. 4 = ( + ) + ( + ) = ln + ln ( )( + ) = ln + + ln + c. + Eample : Evaluate + ( ) ( ) Solution: Let + ( ) ( ) = A + A ( ) + A ( ) + B + = A ( ) ( ) + A ( )( ) + A ( ) + B ( ) + = A ( + )( ) + A ( + ) + A ( ) +B ( + ) = A ( ) + A ( + ) + A ( ) +B ( + ) = A ( ) + A ( + ) + A ( ) +B ( + ) = (A + B ) + ( 4A + A B ) + (5A A + A + B ) +( A + A A B ) Comparing coefficients of powers of we get, A + B = 0, 4A + A B = 0, 5A A + A + B =, A + A A B =, B = A. 4A + A ( A ) = 0 4A + A + A = 0 A + A = 0 A = A. 5A A + A + ( A ) =, A + A A ( A ) =. A + A A =, A + A =. A A =, A + A =.

7 A A = A + A = A = A = A = A = =. A = and B =. + ( ) = ( ) = ln ( ) 7 ( ) + ( )( ) + ln ( )( ) = ln ln ( ) Integration of a rational function containing irreducible quadratic non-repeated factors in the denominator Step - : Assign to each linear factor (a + b) of q() a partial fraction to each non-repeated quadratic factor a + b + c a partial fraction Step - : Consider equation p() q() = A a+b + B+C a +b+c + Step - : Compare coefficients of power of and solve for A, B, C,... Step - 4: Put back values of A, B, C, in step and integrate. Eample : Evaluate Solution: Consider + + ( )( + ) ++ ( )( +) = A + B + C = A( + ) + (B + C)( ) + + = A + A + B B + C C Comparing coefficients we get A + B =, B + C =, A C = A = B B C = i.e. B C = B + C = B + C = + B + C = C = = (A + B) + ( B + C) + A C A a+b and B+C a +b+c.

8 8 C = B = C = = 0 A = B = 0 =. + + ( )( + ) = + + = ln + tan Integration of algebraic rational functions with repeated irreducible quadratic factors in the denominator Step - : If the quadratic factor a + b + c occures n times in the denominator, assign partial fraction A +A a +b+c + A +A A n +A n to it. Assign (a +b+c) (a +b+) n partial fractions + for linear factors. B e+f Step - : Consider the equation p() q() = B e+f + + A +A a +b+c + A +A 4 + (a +b+c) Step - : Compare coefficients of power of and solve for A, A,, B, B, Step - 4: Put back values of A, A,, B, B, in step and integrate. Eample : Evaluate Solution: Consider (+)( +) = A (+)( +) + + B+C + + D+E ( +) = A( + ) + (B + C)( + )( + ) + (D + E)( + ) = A( ) + (B + C)( ) + D + D + E + E = A 4 + A + A + B 4 + B + B + B + C + C + C + C +D + D + E + E By comparing coefficients we get A + B = 0, B + C = 0, A + B + C + D = 0, B + C + D + E =, A + C + E = 0 B = A, B = C A = C A A + A + D = 0 A + D = 0 A + A + D + E = D + E = D = E A + A + E = 0 A + E = 0 A + ( E) = 0 A E = Solving equations A + E = 0, A E =, we get A = 4, E =. C = 4,

9 9 B = 4 D = E = =. ( + )( = + ) ( ) ( + ) = 4 ln ( + ) = 4 ln ( + ) + ( + ) = 4 ln ln + 4 tan + ( + ) + ( + ) = 4 ln ln + 4 tan + ( ) ( () + ) We evaluate I = Put = tan ϕ ( +). = sec ϕ dϕ sec ϕ dϕ sec I = (tan ϕ + ) = ϕ dϕ (sec ϕ) sec ϕ dϕ = (sec ϕ) = sec ϕ dϕ + cos ϕ = cos ϕ dϕ = dϕ = ϕ sin ϕ + = tan tan ϕ + 4 4( + tan ϕ) = tan + +.

10 0 () becomes ( + )( = + ) 4 ln ln + 4 tan tan c = ( ( )) + ln 8 ( + ) + ( ) 4 + c + Eercises [.] Evaluate following integrals. a) ( ), Ans. log + + k b) ( ) +, Ans. + log + + k. c) ( + ( )( 4), Ans. 5 log + d) ( (a b)(c d). Ans. bc ad ln a b c d e) p+q (a b)(c d), Ans. aq+bp [.]Evaluate following integrals. a) ++ (+) (+), a(bc ad) b) + (+) ( ), Ans. 7 ln ) log ( + ) + k. ln(a b) dp+cq Ans. (+) + + log + k (+) ) + k c(bc ad) ln(c d) + k. (+) + k c) , Ans. + ( + ) + 8 ln( + ) 6 ln( + ) + k. d) +, Ans. ln( ) + ( ) ( ) + + ln( ) + k. ( ) [.] Evaluate following integrals. a) ( +5), Ans. 5 ln() 5 ln k. b) ( 4 ( )( +4), Ans. + 0 ln + ) 8 5 tan + k. c) (+) ( +), Ans. ln( + ) 4 ln( + ) ( + ) + k. ( ) ( a) +a + a + k. d) (+a) ( a)( +a ), Ans. ln a [4.] Evaluate following integrals. a) ( 4 +4, Ans. 6 ln (+) + ( ) + ) + 8 [tan ( + ) + tan ( )] + k.

11 b) + +, Ans. (+)( +) 4(+ ) + tan + (+) 8 ln + + k. c), Ans. ( +) ( ) 4 d) + ( )( +), Ans. e) cos (sin +sin ), + + tan + 8 ln (+) ( +)( ) + k tan + 9 ln + + k. Ans. tan (sin ) sin csc + c. (sin +) Integration of some irrational functions Integration of the irrational function (a + b) n. When the epression to be integrated contains n a + b where n Z + and no other irrationality, we put a + b = t n. Eample : Evaluate n a + b. Solution: Put a + b = t n a = nt n dt n a + b = t n a tn dt = n a t n dt = n t n+ a n+ + c = ntn+ (n+)a n+ n(a+b)t n = a(n+). Eample : Evaluate a + b Solution: Put a + b = t b = t dt

12 = t a b, = t b dt t a a + b = t t b b dt = b (t 6 at ) dt = ( ) t 7 b 7 at4 4 = t7 7b a 4b t4 + C (a + b)7/ a(a + b)4/ = 7b 4b = (a + b)7/ 7b a(a + b)4/ 4b = (a + b)4/ (4b a) 8b Formulae: The following formulae will be used: a = +a + a log + +a a + c. a = a a log + a a + c. a = a + a sin a + c. +a = log + +a a + c. a = log + a a + c. a = sin a + c. Integration of irrational functions of the form A+B a +b+c

13 A + B a + b + c = A = A a = A a a + b + c + B a + b b a + b + c + B ( a + b a + b + c + B Ab a = A a a + b + c + ( B Ab a a + b + c a + b + c ) ) a + b + c a + b + c. The last integral can be evaluated by completing the square. Eample : Evaluate Solution: Since +4+ = = (+) = (+) ( ), we have = = 6 = = ( 9) ( ) ( + ) ( ) = log ( + ) + ( + ) = log[( + ) + ( ]. Eample : Evaluate +.

14 4 Solution: Consider + = = ( 4) = + 4 = = = + 4 ( + ( 6) 5 ) 6 ( ) = ( log ( 6) ) + c. Eercises [.] Evaluate following integrals a) +, Ans. + 0 ln ( + 6 b), Ans sin. c) + 6 +, Ans. [ + + ln ( )+ ( d) +, ) (+ 6) ( 5 6) +k. ( ) Ans. + sin. 5 6 ) + 4 ] + k. Intergration of irrational functions of the form (A + B) a + b + c

15 5 Consider (A + B) ( A a + b + c = a + b b + ab ) a a A + b + c A = a (a + b) a + b + c + A ( ) ab a a A b + b + c = A (a + b) a a + b + c ( + B Ab ) a a + b + c = A a ( (a + b + c) / + B Ab ) a a + b + c = A ( a (a + b + c) / + B Ab ) a a + b + c The integral a + b + c can be evaluated by completing the square. Eample. Evaluate ( ) + +.

16 6 Solution: Consider ( ) + + = = ( ) + + ( + 4 ) + + ( ) = ( + ) = ( + + ) / = ( + + ) / = ( + + ) / 7 ( + ) + 4 = ( + + ) / 7 = ( + + ) / log ( + = ( + + ) / 7 4 ( + 6 log ( + ) + ( + ) ( + ) ( ) ( 4 ) ) + ( ( + ) ( + ) + ) ( ) ) 4 ( ) 4 Eample : Evaluate (4 ) + +

17 7 Solution: Consider (4 ) + + = ( 4 ) = 4 ( ) = 4 = 4 = ( + ) + + ) ( ( + ) + + ( ) = ( + + ) / = 4 ( + + ) / ( + 6 = 4 ( + + ) / ( + = 4 ( + + ) / = 4 ( + + ) / + ( ) log ( + ) ) + 9 ( + ) ( ) + ( + ) ( + ) ( ) ( ) ) + (. )

18 8 Eercises [.] Evaluate following integrals a) ( + ) +, Ans. 4 (8 + 0 ) ln ( )+ ( 4 b) + Ans. 4 (8 ) sin 5 + k. ) 4 + k. Reduction Formulae: It will be seen that the method of integrating by parts is essentially a method by successive reduction for, with its help, we are enabled to epress the integral of a product of two functions in terms of another whose evaluation may be simpler. The method of integration by succesive reduction is thus also only a development of the method of integration by parts. Eample : where n Z +. Let I ( +a ) n n =. ( +a ) n I n = ( +a ) n = ( +a ) n

19 9 By integrating by parts, taking as the second function we get, I n = = = = = = = (n )a I n = I n = I n = ( + a ) n ( + a ) n ( + a ( n) ) n ( ( d ) ( a ) n ( n) ( + a ) n ( + a + (n ) ) n ( + a ) n + a a ( + a + (n ) ) n ( + a ) n ( + a + (n ) ) n ( + a ) n + (n )I n (n )a I n ( + a ) n + (n )I n I n ) ( + a ) n ( + a (n )a ) n (n ) (n )a ( + a + ) n (n )a I n n (n )a ( + a + ) n (n )a I n. ( + a ) n Eample : Evaluate Solution: Let I 5 = I n = I 5 = ( +a ) 5. ( +a ) 5. We know reduction formula + n I (n )a ( +a ) n (n )a n + 7 I 8a ( +a ) 4 8a I 6a ( +a ) 6a + I 4a ( +a ) 4a where I 4 = where I = where I = a ( +a ) + a I

20 0 where I = +a = a tan a I 5 = = = = ( + a ) 5 8a ( + a ) [ 8a 6a ( + a ) + 5 [ 6a 4a ( + a ) + [ 4a a ( + a ) + [ ]]]] a a tan a 8a ( + a ) a 4 ( + a ) + 5 9a 6 ( + a ) a 8 ( + a ) a 9 tan a 8a ( + a ) a 4 ( + a ) + 5 9a 6 ( + a ) + 5 8a 8 ( + a ) + 5 8a 9 tan a + c. Eercises [.] Prove the following. a) 0 = 5π (+ ) 4. b) 0 (a + ) 4 = 5π a 7 Reduction Formula for ( + a ) n/. Let I n = (a + ) n/ = (a + ) n/.

21 Using integration by parts taking as second function, we get. I n = (a + ) n/ = (a + ) n/ = (a + ) n/ n = (a + ) n/ n = (a + ) n/ n +na (a + ) n ( ( ) d ) (a + ) n/ n (a + ) n (a + ) n (a + ) n ( + a a ) (a + ) n (a + ) = (a + ) n/ n (a + ) n/ +na (a + ) n = (a + ) n/ ni n + na I n I n ( + n) = (a + ) n/ + na I n I n = n+ (a + ) n/ + na n+ I n. Eample Evaluate ( + a ) 5/. Solution: Consider I 5 = ( + a ) 5/. We know the reduction formula I n = n + (a + ) n/ + na n + I n I 5 = 6 (a + ) 5/ + 5a 6 I where I = 4 (a + ) / + a 4 I

22 where I = ( + a ) / = +a + a log + +a a. I 5 = ( + a ) 5/ = [ 5a 6(a + + ) 5/ 6 4 (a + ) / [ + a + a + a 4 log + ]] + a a = 6 (a + ) 5/ + 5a 4 (a + ) / + 5a4 48 (a + ) / + 5a6 48 log ( + ) + a = 6 (a + ) 5/ + 5a 4 (a + ) / ( ) + 5a4 6 (a + ) / + 5a6 6 log + ( + a ) / a a Eercises [.] Show that a 0 ( + a ) 5/ = 67 Reduction Formula for n Let 48 a6 + 5a. a +b+c 6 ln( + ). I n = n a + b + c a+b b a n = a + b + c = (a + b) n a a + b + c b a = b a I n + a n a + b + c (a + b) n () a + b + c

23 We evaluate the above integral by using integration by parts. (a + b) n [ a + b + c = a + b n a + b + c ] a + b d a + b + c (n ) = n a + b + c a + b + c (n ) n = n a + b + c (n ) n a + b + c = n a a + b + c + b + c (n ) a + b + c n = n [ a n + b + c (n ) a a + b + c +b n a + b + c + c n ] a + b + c () becomes I n = I n (n + ) = = n a + b + c a(n )I n b(n )I n c(n )I n. n a + b + c = b a I n + a [ n a + b + c a(n )I n b(n )I n c(n )I n ] = b a I n + a n a + b + c (n )I n b a (n )I (n )c n I n ) a (n )c I n I n + a a n a + b + c b(n ) b (n )c I n I n an na b(n ) c(n ) I n I n. an an ( b b (n ) a a I n = na n a + b + c + I n = na n a + b + c Eample: Evaluate +.

24 4 Solution: We know the reduction formula I n = n a + b + c = na n a + b + c b(n ) I n an Here n =, a =, b =, c =. Put p() = +. c(n ) I n an I = p() = p() ( )(5) ()() I ()() ()() I = Let n =, a =, b =, c =. I = p() + 5 I 4 I. () p() = p() ( )() ()() I ()() ()() I 0 = + + I I 0 () I = + + = + = = + + I 0 = p() + I 0. ()

25 5 Also I 0 = = + ( ) + = ln[( ) + ( ) + ] + K = ln[( ) + + ] + K. (4) By (), () and (4), () becomes I = 5 p() + 5 p() + = = [ p()] + I I 0 ] 4 I 6 p() I 5 I 0 p() p() = 6 ( ) + [ p() + I0 ] 5 I 0 ln[( ) + + ] + K. Eample: Evaluate I = Solution: We know the reduction formula I n = n a + b + c = na n a + b + c b(n ) I n an c(n ) I n an

26 6 Put p() = + +. Then Now I 4 = 7 p() 4 4 I 9 4 I = 7 [ 5 ] p() p() 4 4 I I 7 p() p() + I + 7 I 7 p() p() + [ p() I ] I I 7 p() p() + = 4 = 4 = I p() + 5 I I 0. I = + + = = = + + I 0 = p() I 0. So Also, I 4 = 4 = 4 I 0 = 7 p() p() + p() [ ] p() I0 I p() p() = 5 7 p() + p() I 0. ( + ) + = ln[( + ) + ( + ) + ] + K = ln[( + ) ] + K.

27 7 Hence I 4 = ( ) + + Eample: Evaluate I = Solution: We know the reduction formula n I n = a + b + c 7 ln[( + ) ] + K.. = na n a + b + c b(n ) I n an c(n ) I n an First we rewrite the given integral as a combination of I 0,..., I and the successively apply the above reduction formula. Also, put p() = Then I = I + 4I 6I + I 0 = p() + 5I + 0 I + 4I 6I + I 0 = p() + 9I 8 I + I 0 = [ 9 p() + 9 p() + I + 5 ] I 0 8 I + I 0 = 9 7 p() p() + 6 I + 5 I and I = = = = I 0 = p() + I 0. So I = = 9 p() p() + 7 p() 9 6 [ p() + I 0 ] + 5 I 0 p() 7 6 p() + 9I0.

28 8 Also, I 0 = = ( ) = sin [( )/ 4] + K. Hence I = 6 ( ) sin [( )/ 4] + K. Eercises: [.] Evaluate ++. Ans. [.] Show that 6 ( 5 + 7) [.] Show that if a 0, then + ln[( + ) ] + K. = 4 ( ) + + a + b + c ln[( + ) ] + K. = 6a (a 5ab + 5b 4ac) a + b + c abc 5b + a a + b + c.

29 9 Formula for π/ 0 sin n, n. Let I n = sin n = sin sin n By integrating by parts we get, I n = sin n sin ( ( ) d sin ) sinn = sin n ( cos ) ( cos )(n ) sin n cos = cos sin n + (n ) sin n cos = cos sin n + (n ) sin n ( sin ) = cos sin n + (n ) (sin n sin n ) = cos sin n + (n ) sin n (n ) sin n = cos sin n + (n )I n (n )I n. I n + (n )I n = cos sin n + (n )I n I n ( + n ) = cos sin n + (n )I n I n = cos sinn n + (n ) n I n.

30 0 For n, consider I n = = = π/ 0 sin n [ cos sin n ] π/ (n ) + I n n 0 n [ cos π sinn π ( cos 0 sin n ) ] 0 + n n n n = (0 + 0) + n n I n. I n = n n I n, I n = n n I n 4, I n = But I = I 0 = Hence I n 4 = n 5 n 4 I n 6, I = I, if n is odd I = I 0, if n is even { n n n n n 5 n n n n π/ 0 π/ 0 π/ 0 Eample: FInd n 4 I, when n is odd n 5 n 4 I 0, when n is even sin = [ cos ] π/ 0 = and = [] π/ 0 = π. sin n = Solution: Consider 0 { n n n n n 5 n n n n. 0. n 4, when n is odd n 5 n 4 π, when n is even I n

31 Put = sin θ = 6 sin θ cos θ dθ, = 0 θ = 0, = θ = π. 0 = Reduction Formula for = = = 8 π/ 0 π/ 0 π/ 0 π/ 0 ( 7 sin 6 ) / θ sin 6 sin θ cos θ dθ θ ( 9 sin 6 ) / θ sin 6 sin θ cos θ dθ θ sin θ 6 sin θ cos θ dθ cos θ sin 4 θ dθ = 8 4 π = 7π 8. π/ 0 cos n, n. Let I n = cos n = cos cos n. By integration by parts, I n = cos n cos ( ( ) d cos ) cosn = cos n (sin ) (sin )(n ) cos n ( sin ) = sin cos n + (n ) sin cos n = sin cos n + (n ) ( cos ) cos n = sin cos n + (n ) cos n (n ) cos n = sin cos n + (n )I n (n )I n I n + (n )I n = sin cos n +(n )I n I n n = sin cos n + (n )I n I n = sin cosn (n ) + I n. n n

32 If I n = π/ 0 cos n then I n = [ sin cos n ] π/ + n n 0 n I n = (n ) (0 0) + I n n I n = (n ) I n n I n = (n ) n I n 4 I n 4 = n 5 n 4 I n 6. I = I if n is odd I = I 0 if n is even I n = But I = I 0 = { n n n n n 5 n n n n π/ 0 π/ 0 I n = n 4 I, n is odd n 5 n 4 I 0, n is even cos = [sin ] π/ 0 = sin π sin 0 = 0 =. = [] π/ 0 = π 0 = π. π/ 0 Eample : Find Solution: Consider cos n = n n π/4 0 π/4 0 = n n n n 5 n n 4 (cos φ) / cos φ dφ., when n is odd n n 5 n n 4 π, when n is even (cos φ) / cos φ dφ. Put sin φ = sin θ, then cos ϕ dϕ = cos θ dθ. φ = 0 θ = 0, ϕ = π 4 θ = π.

33 π/4 0 = π/ (cos ϕ) / cos ϕ dϕ = 0 cos 4 θ dθ = 4 π/ 0 ( sin θ) / cos θ dθ π = π 6. Eercises [.] Evaluate following integrals a) sin 6, Ans. 6 (cos sin5 5 5 cos sin sin + k. b) sin 7, Ans. cos [ sin sin sin + 6 ] 5 + k. c) π/ 0 sin 8, Ans. 5π 56 d) sin 5 8, Ans. 5. e) cos 7, Ans. 7 sin [ cos cos cos f) π/ 0 g) a 0 h) 0 cos 8, / a, (+ ) 5, Ans. 5π 56 Ans. 6 8 πa Ans. 5π 56. ] + k.

34 5. Differential Equations of first order and first degree The construction of Mathematical model to approimate real world problems has been one of the most important aspects of the theoretical development of each of the branches of science. It is often the case that these mathematical models involve an equation in which a function and its derivatives play important roles. Such equations are called differential equations. 5. Functions of two and three variables: Real valued funcions of two variables: Let D = {(, y), y R} be a subset of R and if to each ordered pair there corresponds a single definite value u, where u and ordered pair are associated by some law, then we say that u is a function of two variables and y. We write it as u = f(, y), u is called a dependent variable for f and, y are called independent variables for f. For eample: f(, y) = + y + y. Real valued funcions of three variables: Let D = {(, y, z), y, z R} be a subset of R and if to each ordered triple there corresponds a single definite value u, where u and ordered triple are associated by some law, then we say that u is a function of three variables, y and z. We write it as u = f(, y, z), u is called a dependent variable for f and, y, z are called independent variables for f. For eample: f(, y, z) = z + y + y. Homogeneous functions: A function f(, y) is said to be homogeneous function of degree n, if the sum of the indices of and y in every term is same and is equal to n. Also it can be epressed in the form f(t, ty) = t n f(, y), t > 0. This definition may be etended to functions of three variables. For eample: + y + y, (y + yz + z), sin z ( y + y ), + z are homo- geneous functions of degree,-,0, respectively. Partial Derivatives: Let f(, y) be a function of two independent variables and y defined in a domain D. If y is held constant say at y = b, then f becomes a function of alone and its derivative (if it eists) at = a, is then called the partial derivative at (a, b), f(a, b) of f(, y) with respect to. this is denoted by or f (a, b). It may be

35 denoted simply as f or f.the use of the curly d insted of d suggests the partial derivation. Defination: The partial derivative f at the point (a, b) of the domain,with respect to,is given by f(a, b) f(, b) f(a, b) = lim a a it may be given also as f(a, b) f(a + h, b) f(a, b) = lim h 0 h similarly,if is held constant,say = a, f(, y) becomes a function of y alone and its derivative,(if it eists), at y = bis called the partial derivative of f(, y) at the point (a, b).whenever the menstion of (a, b) is not necessary, it is denoted simply by f y orf y and is given by f f(a, y) f(a, b) y b y = lim y b For eample : If f(, y) = e y, show that f(a, b) = be ab f(a, b) and = ae ab y 5. Differential Equation: An equation f(, y,, d y,, dn y ) = 0 which epresses a relation between dependent and independent variables and their derivatives of any order is called a n differential equation. Further a differential equation which contains only one independent variable is called an ordinary differential equation. For eample:.. + y =. d y + + 5y = 0.. ( + ( ) ) u + u y = y d y =. are eamples of differential equations. u u y + u 7 = 0 are eample of partial differential equation. y Definition: The order of the highest order derivative present in an equation is called the order of the differential equation. Definition: The degree of a differential equation is the highest power of the highest derivative; when all the derivatives in it are cleared of radical signs and

36 fractions. In the differential equation in (), the order is and the degree is. In the differential equation in (), the order is and the degree is. Now consider the differential equation in (), squaring both sides we get, ( + ( ) ) = ( d y ), which is free from radical sign. So, the order is and degree is. Solution of a Differential equaion: A solution of a differential equation is a relation between the variables,which together with the derivatives obtained from it,satisfies the given differential equation. There are two types of solution of a differential equation. (i)general Solution: A general solution of a differential equation is a solution in which the number of arbitrary constants present is equal to the order of the differential equation. (ii)particular Solution: A solution of a differential equation which can be obtained from its general solution by giving particular values to the arbitrary constants in it is called a particular solution. For eample,consider the differential equation d y + = y. It is of order and degree. () Consider the relation y = e () we have, = e = y and d y = e = y. Therefore d y + = y + y = y, therefore relation () satisfy differential equation (). Hence y = e is a solution of equation (). 5. Methods of solving first order and first degree differential equation: The process by which the general solution or primitive is obtained is called solving of the differential equation or method of the solution of the differential equation.there are different methods of solving the differential equation of first order and first degree and they are classified as follows:- (a) Variables seperable form (b) Homogeneous differential equations (c) Non homogeneous differential equations (d) Eact differential equations (e) Linear differential equations (f) Bernoulli s differential equations

37 Variables seperable form : In this type of equation,the variables and y can be seperated so that the coefficient of is a function of alone ( or a constant ) and the coefficient of is a function of y alone ( or a constant ). The solution can be obtained by integrating the equation directly. Let M + N = 0 be a differential equation where M = f() and N = g(y). i.e M is a function of and N is a function of y. such an equation is said to be in variable seperable form. The equation reduces to f() + g(y) = 0. Integrating the equation, we get the general solution as f() + g(y) = c where c is an arbitrary constant. ILLUSTRATIVE EXAMPLES :- Eample. Solve the differential equation ( + ) y = 0 Solution:- It can be written as y = Integrating. (variable seperable form ) + y = + + c Therefore log(y) + + = c log(y) ( + ) = c log(y) + log( + ) = c log y( + ) = + c is required general solution. Eample. Solve y = y + ( + ) Solution :- It can be put as y y = ( + ) Therefore + = y y variables are seperated,therefore integrating + = y( y) + c

38 log( + ) = ( A y + B y ) + c log( + ) = ( y + ) + c where A = andb = y log( + ) = log(y) + log( y ) + c log( + ) = log(y) log( y) + c log( + ) + log( y) log(y) = log(k) log( ( + )( y) ) = log(k) y ( + )( y) = yk, is required general solution. Eample. Solve = y + e y/ Solution :- Given differential equation is = y + e y/ () put y = v, therefore = v + dv Therefore () takes the form. () (v + dv ) = v + e v (v + dv ) = v + e v dv = e v dv e v = Therefor integrating, (variable seperable form) dv e v = + c e v dv = log() + c e v = log() + c e y/ = log() + c, is required solution.

39 Eample 4. Solve (sin + cos ) + (cos sin ) = 0 Solution :- Given equation can be written as, (cos sin ) = (sin + cos ) Integrating, cos sin = sin + cos + c y = log(sin + cos ) + c y + log(sin + cos ) = c, is required general solution. Eample 5. Solve + b y a = 0 Solution :- The given equation can be written as, = b y a b y = a Integrating, b y + a = c (variable seperable form) (variable seperable form) Therefore sin (y/b) + sin (/a) = c, is general solution. Eercise 5. Solve the following differential equations..sin + cos = 0..sec tan y + sec y tan = 0.y + e +y = 0. 4.( + y ) =. 5.( y)e y/ + y( + y ey/ ) = = (4 + y).

40 7. + esin = y = = e+y + e y. 0. = tan ( + y). Answers 5.. tan tan y = c.. tan tan y = c. ( + y) = e y (c + e ) 4. tan y + = c 5. e y/ + y = c 6. (4 + y) = tan( + c) 7. y + e sin = c 8. = ce sin y 9. e + e y + = c 0. y + sin( + y) + cos( + y) = c.solve( + y) = a..solve = e +y + e y. Problems for practical.solve( + y ) tan + y( + ) = 0. 4.Solvee tan y + ( e ) sec y = 0. 5.Solve cosec log(y) + y = 0. Homogeneous Differential Equations :- Any function f(, y) is said to be homogeneous function of n th degree if, f(t, ty) = t n f(, y) Consider the function, f(, y) = 5y + y Therefore f(t, ty) = t 5tty + t y

41 = t ( 5y + y ) = t f(, y) Thus 5y + y is homogeneous function of degree in and y. Definition: The differential equation M+N = 0 is said to be homogeneous differential equation if both M and N are homogeneous function of same degree in and y. OR : The first order differential equation of the type f(, y) = ;where f(, y) g(, y) and g(, y) are homogeneous function of the same degree. It is called homogeneous type differential equation. Methods of solving the homogeneous differential equation :- We have f(, y) = g(, y) As f and g are homogeneous function of the same degree say n; () f(, y) = n φ( y ) and g(, y) = n ψ( y ) Therefore () can be put as; = n φ( y ) n ψ( y ) = φ( y ) ψ( y ) = ψ( y ) () put y = v y = v = v + dv Therefore equation () reduces to v + dv = H(v) dv = H(v) v dv H(v) v = Integrating this,we get dv H(v) v = log() + c The back substitution v = y u gives the general solution. (variable seperable form)

42 ILLUSTRATIVE EXAMPLES Eample. Solve y ( + y ) = 0 Solution:- Given differential equation is y ( + y ) = 0 y = ( + y ) = y + y It is homogeneous type differential equation,since y and + y are homogeneous function of the same degree that is two. () put y = v ; therefore = v + dv Therefore () becomes, v + dv = v + v dv = v + v v dv = v v v + v = v + v + v v dv = variables are separated,integrating () () + v v dv + = c ( v + )dv + log() = c v log(v) v + log() = c log(v) = v + log(k) log( y k ) = y y k = /y e, is required general solution. Eample. Solve y + = y c = log(k) from()

43 Solution:- This is a homogeneous equation of degree and hence we put y = v then = v + dv The equation reduces to v + (v + dv ) = v (v + dv ) v + v + dv = v + v dv by cancelling throughout v + dv dv = v v = (v ) dv = v = ( v v )dv Integrating = dv v dv + c log() = v log(v) + c log(v) = v + c Therefore,. y = e(y/)+c y = e (y/)+c, is required general solution. Eample. Solve ( + y ) = y Solution :- This is a homgeneous of degree.hence we put y = v then, = v + dv ( + v )(v + dv ) = v ( + v )(v + dv ) = v ( + v )(v + dv ) = v v + v 4 + ( + v ) dv = v v 4 + ( + v ) dv = 0

44 v 4 = ( + v ) dv + v = v 4 dv = ( v 4 v )dv dv dv = v 4 v log() = log(v) log(c) v log() + log(v) + log(c) = v log(vc) = v log yc = y. Therefore, cy = e /y, is general solution. Eample 4. Solve the differential equation y + sin y = 0 Given differential equation is y + sin y It is homogeneous. Hence,put y = v, therefore = v + dv () becomes v + sin( v dv ) (v + ) = 0 v + sin v v dv = 0 dv sin v = 0 () Variables are separated,integrating dv sin v = c cosecvdv log() = c () () log(cosecv cot v) log() = log(k) cosecv cot v = k cosec y cot y = k, is general solution.

45 Eample 5. Solve ( tan y y sec y ) + sec y = 0 Solution :- The given equation can be written as, ( tan y y y sec ) + y sec = 0 put y = v then = v + dv ( tan v v sec v) + sec v(v + dv ) = 0 tan v + sec v dv = 0 sec v tan v dv = sec v Integrating, tan v dv = + log(c) log(tan v) = log() + log(c) = log c tan v = c Therefore, tan y = c is general solution. Eercise 5. Solve the following differential equations.. ( + y ) y = 0. = y + y 5. = y + cos y. = y + y 4. y = ( + y ) 6. = + y y( + y) 7. ( + y) y = 0 8. (4y ) + 4( + y) = 0 9. y cos y [ y sin( y ) + cos( y )] = 0 0. = + y y Answers

46 . y = c. ( y) = cy. yy = c 4.sec y + log(y) = c 5. tan y = clog() 6.y = + ce y/ 7. c y = ey/ 8. 4y + 8y = c 9. y sin( y ) = c 0. tan ( y ) = log( + y ) + c Problems for practical. Solve the following differential equations. 4 y. = y. y = y 5. y = ylog( y ). (y sin( y ) + ) sin( y ) = 0 4. ( + e /y ) + ( )e/y y = 0 Non Homogeneous Differential Equations (Differential equation reducible to homogeneous form) Definition : The differential of the type = a + b y + c is called the non-homogeneous differential equation;where a + b y + c a, b, c, a, b, c are constant and both c, c are not zero simulteneously. Method of solution : Consider a non homogeneous linear differential equation = a + b y + c a + b y + c () CaseI : Suppose a = a = 0. Then equation () becomes = b y + c = f(y) b y + c Therefore = f(y) which when integrated, will give a solution for the given differential equation. CaseII : Suppose b = b = 0, then similarly equation () will become = g() which is again in the variable separable form and will hence give a solution for the differential equation.

47 CaseIII : If a a = b b, then we can write a a = b b = k (say) i.e a = ka ; b = kb, therefore () becomes = a + b y + c = a + b y + c ka + kb y + c k(a + b y) + c Now, put u = a + b y, then du = a + b therefore () gives du a = u + c = F (u) b ku + c Further simplification yields du a + b F (u) = since this is in the variables separable form, we can solve it as usual. CaseIV : If a a b b we put = X + h, y = Y + k where h and k are constants to be suitably determined.therefore equation () takes the form. dy dx = a X + b Y + (a h + b k + c ) a X + b Y + (a h + b k + c ) Net choose h and k, so that a h + b k + c = 0 and a h + b k + c = 0 () Note that a b a b 0, since a a b b Hence h and k can be determined by solving the equations (). With this choice of h and k, the given differential equation is transformed to dy dx = a X + b Y a X + b Y This is a homogeneous differential equation and a solution of () can thus be obtained. ILLUSTRATIVE EXAMPLES. Eample. Solve 6 y 7 = + y 6 Solution : Here a = 6, b =, c = 7 a =, b =, c = 6 a =, b = a b Therefore a a b b Hence put = X + h and y = Y + k Therefore = dy dx

48 Given equation becomes dy 6(X + h) (Y + k) 7 = dx (X + h) + (Y + k) 6 dy 6X Y + (6h k 7) = dx X + Y + (h + k 6) we choose h and k in such a manner that 6h k 7 = 0 and h + k 6 = 0 i.e h = and k = dy 6X Y = which is homogeneous dx X + Y Hence put Y = V X, therefore dy dx = V + X dv dx V + X dv dx 6X V X = X + V X = 6 V + V X dv 6 V V V = dx + V V + V + 4V 6 = dx X V + dx Integrating V + 4V 6 dv = X + log(c) log(v + 4V 6) + log(x) = log(c) log[(v + 4V 6)X ] = log(c). (variable separable form) (V + 4V 6)X = C Y X + 4 Y X 6 = C X Y + 4XY 6X = C (y ) + 4( )(y ) 6( ) = C y 6y + + 4y 4 6y = C y + 4y y = C y + 4y y = C + 9 y + 4y y = k This is general solution.

49 Eample. Solve = y + y + 5 Solution : We have, = y + y + 5 = y + ( + y) + 5 Here, a a = b b = Therefore put y = v, therefore = dv = dv from () dv = v + v + 5 dv = v + v + 5 = v + 5 v = v + v + 5 v + 5 v + 5 dv = v + Integrating, v + 5 v + dv = + c (v + ) ( v + + v + )dv = + c dv dv + v + = + c v + log(v + ) = + c () (variable separable form) ( y) + log( y + ) = + c is the required solution. Eample. = + y + + y Solution : Here, a a = b b = put + y = v () + = dv Equation () becomes, dv = v + v dv = + v + v = v + v + v

50 dv = v v v v dv = which is variable separable form. Integrating, ( v )dv = + c v log(v) = + c + y log( + y) = + c y + log( + y) = c Eample 4. Solve (7 y + ) = (7y 7) Solution : we can put it as 7 y + = + 7y 7 Here, a = 7 a, b = b 7 Therefore, a a b b Put = X + h and y = Y + k, therefore = dy dx () becomes dy dx we take 7h k + = 0 h + 7k 7 = 0 solving these equations; we get h 0 = k 40 = 40 Therefore () gives dy dx = = 7X Y + (7h k + ) X + 7Y + ( h + 7k 7) therefore h = 0, k = 7X Y X + 7Y which is homogeneous, put Y = V X () () () (4) dy dx = V + X dv dx therefore (4) becomes V + X dv dx = X dv dx = 7 V + 7V V = 7V 7 7V 7X V X X + 7V X V 7 V dv = dx X (variable separable form)

51 V Integrating 7 dx V = X + C V V dv 7 V dv = log(x) + C log(v ) 7 ( ) log(v V + ) + log(x) = C log[ (V )X 4 ] = log(k) ( V V + ) 4(V ) 7 X 4 (V + ) (V ) = k X 4 (V + ) 0 (V ) 4 = k X 4 ( Y X + )0 ( Y X )4 = k X 4 (Y + X)0 + (Y X) 4 X 4 = k Therefore (y + + ) 0 (y ) 4 = k is required general solution.. Eercise 5. Solve the following differential equations. = + y + + y +. ( + y ) + ( y + 4) ( 5y + ) = ( + 4y 6) 6. ( + 4y ) + ( + y + ) = 0. ( + y ) = (6 + y + ) = + y y + Answers. y + log( + y) = c. 7( y) = 5log(4 + 7y + ) + c. y + y 4 + 8y = c 4. y y 0 + y = c 5. ( 4y + )( + y ) = c

52 6. + y + y + y = c Problems for practical Solve the following differential equations.. ( + 4y + ) = ( + y + ). ( y ) ( y ) = 0. ( + y ) ( y + ) = 0 4. = + y + y Eact differential equation:- Definition : If the differential equation is obtained only differentiating a relation of the type u(, y) = c without any operation of elimination, multiplication or division then that differential equation is called Eact Differential Equation. Consider the relation u(, y) = y = c () since it contains only one arbitrary constant c, differentiating () w.r.to + y = 0 () () is differential equation of (). since it is obtained only differentiating () without any elimination, multiplication or division, it is eact differential equation. Theorem : To determine the necessary and sufficient condition for differential equation of first order and first degree to be eact. Statement : The necessary and sufficient condition for the differential equation M + N = 0 () to be eact is M y = N Proof : Necessary part :- Suppose the equation M+N = 0 is eact equation. Therefore for some function u = u(, y), we have u u = M and y = N u y = M y and u y = N But the second order partial derivatives commute, M y = N ()

53 Sufficient part :- Suppose M y = N Let φ be a function such that φ = M y φ = M y φ y = N φ y = N φ y = φ y ) ( φ y ) = N (since () by() φ y = φ y ) Integrating both sides of this equation w..r.t., holding y as a constant and noting that the constant of integration may be a function of y, say ψ(y) Therefore φ y = N + ψ(y). Now consider the function u(, y) defined as below. u(, y) = φ(, y) ψ(y) for this function u, we have u = φ [ ψ(y)] u = φ 0 u = M (5) Also u = φ y y [ ψ(y)] u y = φ y ψ(y) u = N by (5) (6) y M + N = u u + y Therefore equation M + N = 0 is an eact differential equation. (4)

54 ILLUSTRATIVE EXAMPLES. Working rule for the general solution of M + N = 0 which is eact. i) Integrate M with respect to keeping y constant. ii)integrate N with respect to y, whose terms are free from. iii) Add result (i) and (ii) and equate it to an arbitrary constant. Mathematically it can be put as M + (Terms in N free from ) = c y as a constant Eample. Solve ( cos y + y) + ( sin y + y) = 0 Solution : The given differential equation is ( cos y + y) + ( sin y + y) = 0 Here, M = cos y + y and N = sin y + y M y = sin y + and N = sin y M y = N Thereore, the given differential equation is eact. Integrated M w.r.t., keeping y as a constant M = ( cos y + y) = cos y + y = cos y + y = cos y + y () Now integrate N w.r.t. y, those terms are free from y = y = y () Adding () and () and equate i.e an arbitrary constant

55 cos y + y + y = c Eample. Solve cos y sin y = sec Solution : Her M = cos y sec and N = sin y M N = sin y and y = sin y M y = N Therefore the given equation is eact. General solution of given equation. (cos y sec ) + (0) = c yconstant cos y tan = c is general solution. Eample. Solve [y( + ) + cos y] + [ + log() sin y] = 0 Solution : Here M = y( + ) + cos y and N = + log() sin y M y = ( + N ) sin y and = + sin y M y = N Therefore it is eact differential equation. Hence general solution is given by [y( + ) + cos y] + 0 = c y constant y + ylog() + cos y = c Eample 4. Solve ( + e /y ) + e /y ( y ) = 0 Solution : Here M = + e /y and N = e /y ( y )

56 M y = e/y y and N y = e/y ( y ) + ( y )e/y ( y ) = y e y M y = N y Therefore, it is eact. Hence general solution is ( + e y) + o = c y constant + e/y /y = c + ye/y = c is general solution. Eample 5. Solve (a + hy + g) + (h + by + f) = 0 Solution : Here M = a + hy + g and N = h + by + f M y = h = N y Therefore it is eact. M = (a + hy + g) = a + hy + g () and N = (by + f) = b y general equation is () + () a + by + hy + g + fy = c a + by + hy + g + fy = c + fy () Eercise 5.4 Show that the following differential equations are eact and solve them.. [ + y cos(y)] + cos(y) = 0. y + = + y + 4. (y ) + ( + y) = 0

57 4. (sec. tan tan y e ) + sec sec y = 0 5.( ay) = (a y ) 6. = y + y 5 Answers.. sin(y) = c. ( y ) 4y + 8y = c. y + y = c 4. sec tan y e = c 5. ay + y = c 6. y + y + 5y = c Problems for practical. Show that the following differential equation are eact and solve them.. [y( ) sin y] + [ log() cos y] = 0. ( y y + ) ( y + ) = 0. (cos y y ) + (cos y sin y y) = 0 4. (cos cos y cot ) sin sin y = 0 5. (sin + cos y ) + (y cos + sin y + y) = 0 Integrating factors Definition : If a differential equation which is not eact becomes eact when it is multiplied by some function of and y say µ(, y), then µ is called an integrating factor (I.F) of the given differential equation. Remark: Finding an integrating factor by inspection, depends largely upon recongnisation of certain common eact differentials. In this connection the following list of eact differential should be noted carefully. d( y y ) =. d( y ) = y y

58 . d(y) = + y 4. d( y ) = y y 5. d(tan y ) = y + y Rule: If M + N = 0 is a homogeneous differential equation then is an integrating factor, provided M + Ny 0 M + Ny Proof : Consider M + N = 0 () Let M and N be homogenous functions of and y of degree say n. Suppose M + Ny 0.Then multiply equation () by to get M + Ny M M + Ny + N M + Ny = 0 i.e M + N = 0 () where M = M M + Ny, N = N M + Ny we show that () is an eact differential equation. For this we shall make use of Euler s theorem for homogeneous function which is stated as follows. If u is a homogeneous function of and y of degree n then du + y du = nu. By applying Euler s theorem for both M and N we have, M + y M y = nm N + y N y = nn consider, M y Ny M M (M + Ny) y = My N ( ) MN M( M y + y N y + N) (M + Ny) y = (M + Ny) M and N (M + Ny) = N N( + M + y N ) (M + Ny) Thus, M y N = M N N M MN (M + Ny) = N[y M y = N(nM) M(nN) (M+Ny) + M N ] M[y y + N ] (M + Ny) by(iii)

59 i.e M y = N = 0 Therefore, equation () is eact. Hence M + Ny Eample. Solve ( y y ) ( y) = 0 is an integrating factor. Solution : The given differential equation is ( y y ) ( y) = 0 () clearly () is a homogeneous differential equation comparing () with M+N = 0, we have M = y y and N = ( y) M + Ny = ( y y ) y( y) = y 0 I.f of () = M + Ny = y on multiplying () by, we have y ( y ) ( y y) = 0 which must now be an eact equation so,the required solution by the ususal rule is log() + log(y) = c y Eample. Solve y ( + y ) = 0 Solution : The given differential equation is y ( + y ) = 0 () Here, M = y and N = y M y = = N It is not eact. M and N are homogeneous function of the same degree that is and M + Ny = y ( + y )y = y 4 0 M + Ny = is I.F y4 on multiplying () by y 4

60 ( y y 4 + y y 4 = 0 ( + ( + y) = 0 () y y4 solution of () is y + y = c + log(y) = c is general solution y Rule : If the equation M + N = 0 is of the type f (y)y + f (y) = 0 is an I.f. provided M + Ny 0 then M Ny Proof : Here M = yf (y) and N = f (y) suppose M Ny 0, then M Ny = y(f f ) multiplying the given equation by () where M = Then, M y N = M y f (f f ) and N = Ny we obtained the equation M +N = 0 f y(f f ) = (f f )f y (y) f (f f ). y (y) (f f ) = (f f f f f f + f f ) (f f ) = f f f f (f f ) () is eact differential equation; which gives M Ny Eample. Solve ( y + 4y + ) + ( y + 5y + )y = 0 Solution : The equation is in the form f (y)y + f (y) = 0 Here M = f (y)y and N = f (y), then M Ny = y = 0 y is an I.F, multiplying the given equation with y gives is an integrating factor.

61 ( + 5 y + 4 )y + ( + y y + ) = 0 y This is an eact differential equation. Therefore solution is y + 5log() y + 4log(y) = c is the solution. (y y ) + 4 y = c Eample. Solve (y sin y + cos y)y + (y sin y cos y) = 0 Solution : The given differential equation is in the form f (y)y + f (y) = 0 Here M = y(y sin y + cos y) and N = (y sin y cos y) M Ny = y((y sin y + cos y) y((y sin y cos y) = y cos y 0 Hence M Ny = y cos y is an I.F multiply the given differential equation by the I.F on simplification, we get (tan y + )y + (tan y ) = 0 y y This is an eact differential equation. Hence solution is (y tan y + y ) y = log(c) i.e log(sec y) + log() log(y) = log(c) i.e sec y = cy is the solution of the given differential equation. M y N Rule : If is a function of alone say f(), then e f() is an I.F of the M equation M + N = 0 Proof : Suppose M y N M = f() () then multiplying M + N = 0 by e f(), we get M + N = 0, where M = Me f() and N = Ne f()

62 consider, M y = e f(). M y and N = e f(). N + N [e f() ] M y N = e f() [ M y N Nf()] by () e f() is an I.f N M y = e f() [ N + Nf()] = e f() [Nf() Nf()] = 0 Rule4 : If is a function of y alone say g(y), then e (g(y)) is an I.f of the M differential equation M + N = 0 Suppose N M y M = g(y) () then multiplying M + N = 0 by e (g(y)), we get M + N = 0, where M = Me g(y) and N = Ne g(y) M y = e g(y). M y + M y (e g(y) = e g(y) [ M y + Mg(y)] N = e g(y). N N M y = e g(y) [ N M y Mg(y)] e g(y) is an I.F = e g(y) [Mg(y) Mg(y)] by () = 0 Eample. solve ( + y + ) + y = 0 Solution : Given ( + y + ) + y = 0 ()

63 compairing this equation with M + N = 0, we have M = + y +, N = y M y = y, N = y Therefore M y N. given equation is not eact N ( M y N ) = y (y y) = which is function of alone I.F of () is e f() = e (/) = e log() = multiply both the side of equation () by ( + y + ) + y = 0 which must be eact equation and so its solution is y + = c or y + 4 = c Eample. Solve (y 4 e y + y + y) + ( y 4 e y y ) = 0 Solution : Compairing the given equation with M + N = 0, we have M = y 4 + y + y ; N = y 4 e y y. M y = 8y + 6y + ; N = y4 e y y N M y = 4(y e y + y + ) 4 y (y4 e y + y + y) = 4, which is a function of y alone. y I.F = e f(y) = e ( 4/y) = e 4log(y) = y 4 multiplying given equation by, we have y4 (e y + y + y ) + ( e y y y 4 = 0

64 which must be eact and so by usual method its solution is e y + y + y = c Eample. Solve ( + y + ) + y = 0 Solution : Here M = + y + and N = y M y M y = y and N = 0 N N = y y = since the R.H.S contains no terms in y, we conclude that of alone. e = e is an I.F M y N N = is a function multiplying the givn equation by e, we obtain as eact differential equation. Thus e ( + y + ) + e y = 0 is eact solution is M + 0 y constant e ( + y + ) = c y constant e e. + y e + e. = c i.e ( + y )e = c is required solution Eample 4. solve y( + y + ) + ( + y + ) = 0 Solution : since M = y + y + y and N = + y + Then N M y M = which is a function of y alone. + y + y y( + y + = + y + y( + y + = y I.F = e y = e logy = y multiplying the given equation by y, we get eact differential equation.

65 y ( + y + ) + y( + y + ) = 0 solution is (y + y + y ) + 0 = c y constant i.e y + y + y = c Eercise 5.4 Solve the following differential equations.. (y + y) + ( y + + y 4 ) = 0. ( y 4 + y) + ( y ) = 0. = + y y 4. y( + y + ) + (y 4 ) = 0 5. (y + y y + 6) + ( + y ) = 0 6. (8y 9y + ) + ( y) = 0 7. ( + y ) = y 8. (4y + y ) + ( + y) = 0 9. (y 4 + y) + (y + y 4 4) = y( y) + ( y) = 0 Answers. y 4 + 6y + y = c. y + = cy. log( + y ) + tan ( y ) = c 4. y y + y + y = c 5. y + y y + = c

66 6. 4 y y + 4 = c 7. = y + cy 8. (4y + 4y ) = c 9. y + y + y = c 0. y( y) = c Problems for practical.. (y + ) + y (y + 4) = 0. y(y + y ) + (y y ) = 0. (y + y) + ( y) = 0 4. ( y y ) ( y) = 0 5. (y + y) (y + ) = 0 6. ( + y + y) + ( + ) = 0 7. ( + y + ) y = 0 Linear Differential Equation Definition : The differential equation of the type + P y = Q, where P and Q are function of only, are called linear differential equations of first order and first degree or linear in y.. + y tan = log(). + y + = e.. log( + )y = e cot are eamples of linear differential equation.

67 Method of solving the linear differential equation : + P y = Q Consider the linear differential equation of first order. [Q() P ()y] = 0 for this equation we have, M = Q() P ()y and N = + P ()y = Q() () N ( M y N ) = ( P () 0) = P () which is a function of alone. Integrating factors of the above equation is e P () e P (). + e P () [Q() P ()y] = 0 is eact different equation. Its solution is ye P () + Q()e P (). + c = 0 ye P () = Q()e P (). + c This is the general solution of (). Note. In some differential equations, by taking suitable substitutions, they can be reduced to linear form. Note. If is dependent variable and y is independent variable then the linear equation is + P (y) = Q(y). Eample. Solve + cot y sin = sin cosecy Solution : It can be put as cosecy. + cos y. sin = sin sin y cosecy sin y + cos y sin = sin () put cos y = v, sin y = dv () Hence () becomes

68 dv + v sin = sin dv v sin = sin () () is linear in v Here P = sin, Q = sin and I.F = e P = e sin = e cos Hence, general solution of () is v.e cos = e cos ( sin ) + c put cos = t sin = dt ve cos = e t.dt + c ve cos = e t + c ve cos = e cos + c cos y e cos = e cos + c is general solution. Eample. Solve (y + y) + ( + ) = 4 Solution : The given differential equation can be written in the form + ( + ) y = 4 ( + ) Here P = + and Q = 4 ( + ) I.F is e p = e so that a general solution is y.e p = e P.Q + c i.e y ( + ) = ( + ) + c + = e log(+) = ( + ) Eample. Solve cos + y ( sin + cos ) = Solution : The given equation can be written as

69 + (tan + sec )y = (tan + I.F = e ) = e log(sec )+log() = e log(). sec = sec Hence the required solution is y. sec = sec. + c i.e y. sec = tan + c Eample 4. Solve ( )( ) + y = Solution : The given equation is + y = Here P =, Q = P = = log( ) so I.F = e P = so the required solution is y =. = e / dt + c (put = t = dt) or y = + c = e / + c Bernoulli s Equation : The D.E of the type + p()y = q()yn () is called Bernoulli s equation. The presence of the term q()y n suggests that it is not linear equation. It can be reduced to the linear form by using the substitution y n = v Method of solving the Bernoulli s equation : Substituting v = y n in (), we have ( n)y n = dv

70 = dv yn n D.E () takes the form dv yn + p()y = q()yn n dv + ( n)p() y n = ( n)q() dv + ( n)p() v = ( n)q() This is the linear equation in v and. The integrating factor is I = e ( n)p() and the solution is v.i = ( n)q()i + c y n e ( n)p() = ( n)q() e ( n)p() + c Eample. Solve + y + y4 = 0 Solution : The given differential equation is a Bernoulli s equation as it can be written in the form + P y = Qyn, where p =, Q =, n = 4 dividing by y 4 to given equation y 4 + y = put y = u y 4 = du du 6u = This is linear differential equation with P = 6 and Q = I.F is e p = e 6 = e Therefore, solution is ue = e + c

71 which gives the general solution y + = ce Eample. ( ) + y log(y) = ye Solution : Dividing by y, the given equation reduces to y Let log(y) = v so that y = dv + log(y) = e () using (),() gives dv + v = e () compairing () with dv + P v = Q, we have () P = and Q = e I.F = e P = e log() = Hence solution of () is v. = Q.. + c v =.e. + c log(y) = e e + c Eample. Solve + y cos = yn sin Solution : the given differential equation is + y cos = yn sin () This is Bernoulli s equation put y n = v ( n)y n = dv = dv yn n Equation () reduces to dv yn n + y cos = yn sin

72 dv + ( n)y n cos = ( n) sin dv + [( n) cos ]v = ( n) sin This is linear equation in v and. Here p() = ( n) cos and q() = ( n) sin The integrating factors is I = e p = e ( n) cos = e ( n) sin Therefore, solution of equation () is v.i = q()i + c y n e ( n) sin = ( n) sin e ( n) sin + c put sin = t on R.H.S Therefore, cos = dt y n e ( n) sin = ( n) t e ( n) t dt + c Integrating by parts y n e ( n) sin = ( n)[t e( n)t n frace ( n)t n] + c y n e ( n) sin = ( n)[ n sin e( n) sin e( n) sin ( n) y n e ( n) sin = [sin e ( n) sin n e( n) sin ] + c y n e ( n) sin = e ( n) sin [sin n ] + c This is the general solution. Eercise 5.5 Solve the following equations:. y ( + y + ) = 0. ( + y ) + y = sin y = cos y + y = cos 4. sin ( ) + y = cos 6. cos + y = tan