NATIONAL ACADEMY DHARMAPURI TRB MATHEMATICS DIFFERENTAL EQUATIONS. Material Available with Question papers CONTACT ,

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1 NATIONAL ACADEMY DHARMAPURI TRB MATHEMATICS DIFFERENTAL EQUATIONS Material Available with Question papers CONTACT , TEST BACTH SATURDAY & SUNDAY NATIONAL ACADEMY DHARMAPURI

2 Unit-VIII - Differential Equations Linear differential equation - constant co-efficients - Existence of solutions Wrongskian - independence of solutions - Initial value problems for second order equations - Integration in series - Bessel s equation - Legendre and Hermite Polynomials - elementary properties - Total differential equations - first order partial differential equation - Charpits method DIFFERENTIAL EQUATIONS Definition: An equation involving one dependent variable and its derivatives with respect to one or more independent variables is called a Differential Equation. Differential equation are of two types. (i) Ordinary Differential equation (ii) Partial Differential equation Ordinary Differential equation An ordinary differential equation is a differential equation in which a single independent variable enters either explicitly or implicitly. Partial Differential equation A partial differential equation is one ine which at least two independent variable occur. Example : x z x +y z y = z Order The order of a differential equation is the order of the highest differential coefficient degree The degree of the differential equation is the degree of the highest order derivative after removing radicals and fractions Problems Find the oder and degree of the differential equation 1. [1 + ] 3 = ( d y order =, degree =. cosx d y +sinx ) +8y=tanx

3 order =, degree = 1 3. d y =0 order =, degree = 4. d y =[4 + 3 ] 4 order =, degree = 4 5. d y +x = y + order = 1, degree = 6. (1 + y ) = y order = 1, degree = 1 7. y =4 +3x order = 1, degree = 8. sinx(+) = cosx(-) order = 1, degree = 1 9. d 3 y y 3+(d ) y = 7 order = 3, degree = d 4 x x dt 4+d dt + dt order = 4, degree = 1 5 = e t (TRB) Formation of differential equations : Let f (x, y, c 1 ) = 0 be an equation containing x, y and one arbitrary constant c 1. If c 1 is eliminated by differentiating f (x, y, c 1 ) = 0 with respect to the independent variable once, we get a relation involving x, y and If we have an equation f(x, y, c 1, c ) = 0 containing two arbitrary constants c1 and c, then by differentiating this twice, we get three equations (including f). If the two arbitrary constants c 1 and c are eliminated from these equations, we get a differential equation of second order. Problems Form the differential equation by eliminating arbitrary constant 1. y = ax+a Differentiating with respect x = a

4 differential equation is, y = x + dt. y = Acosx+Bsinx Differentiating with respect x = -Asinx+Bcosx Again Differentiating with respect x d y d y = -Acosx-Bsinx = -(Acosx+Bsinx) =-y +y = 0 3. y = Acoskx+Bsinkx Differentiating with respect x = -Aksinx+Bkcosx Again Differentiating with respect x d y = -Ak cosx-bk sinx = -k (Acosx+Bsinx) =-k y d y +k y = 0 4. y = Acos4x+Bsin4x Differentiating with respect x = -4Asinx+4Bcosx Again Differentiating with respect x d y d y = -16Acosx-16Bsinx = -16(Acosx+Bsinx) =-y +16y = 0 5. y =Ax +Bx+C Differentiating with respect x yy = Ax+B Again Differentiating with respect x [yy +y y ] = A Again Differentiating with respect x

5 yy +3y y +y y = 0 yy +3y y = 0 (or) y d 3 y 3+3d y =0 6. y=ae 3x + Be 3x y = 3Ae 3x 3Be 3x y = 9Ae 3x + 9Be 3x = 9(Ae 3x + Be 3x ) y =9y y -9y = 0 7. y=ae 3x + Be 5x (1) y = 3Ae 3x + 5Be 5x () y = 9Ae 3x + 5Be 5x (3) From equation (1),() and (3) eliminating A,B y 1 1 y 3 5 = 0 y 9 5 y -8y +15 = 0 (OR) y = Ae p1x + Be px is a solution of (D-p 1 )(D-p ) y= 0 D 3 D 5 y = 0 (D -8D+15)y = 0 (or) d y 8. y = c 1 e x + c e x (TRB) (D-1)(D+)y = 0 (D -D+)y=0 d y 9. y=ae x + Be x (D+)(D-)y = 0 (D +4)y=0 d y - +y = 0 4y = y = e 3x (Ccosx+Dsinx) ye 3x = Ccosx+Dsinx (1) Differentiating with respect x y(-3e 3x )+ e 3x y =-Csinx+Dcosx e 3x [-3y+y ] = -Csinx+Dcosx Again Differentiating with respect x y = 0 e 3x (-3y +y )-3(3y+y ) e 3x = -4Ccosx-4Dsinx = -4 (Ccosx+Dsinx) e 3x [y -6y +9y] = -4ye 3x y -6y +13y = 0 (OR) If p = α ± iβ,then Differential equation, (D -(sum of roots)d+product of roots )y = 0

6 (D -(3+i+3-i)D+(3+i)(3-i))y = 0 (D -6D+13 )y = y = (Ax+B)e 3x Differential equation is,(d-3)(d-3) y = 0 (D -9)y=0 Solutions of first order and first degree equations: (i) Variable separable method (ii) Homogeneous differential equation (iii) Linear differential equation Variable separable : Variables of a differential equation are to be rearranged in the form f 1 (x) g (y) + f (x) g 1 (y) = 0 i.e., the equation can be written as f (x)g 1 (y) = f 1 (x) g (y) g 1 (y) = f 1 (x) and integrating on both sides g (y) f (x) 1. +(1 y integrating g 1 (y) = f 1(x) g (y) 1 x )1 = 0 = 1 y 1 x = 1 y sin 1 y sin 1 x = c. Solve = y x 3 y = x 3 integrating 3. Solve - 1 y = x 44 +c = y+ x 1 f (x) y = x3 1 x

7 y+ integrating = x 1 y+ = x 1 log(y+) = log(x-1)+logc log( y+ ) = logc x 1 y+ x 1 = c 4. Solve e x tany+(1-e x )sec y = 0 e x = 1 e x -sec y tany integrating e x = sec y 1 e x tany -log(1- e x ) = - logtany+logc log( tany 1 e tany 1 e x ) = logc x = logc 5. Solve x 1 + y +y 1 + x x = - y 1+x integrating 1+y x = y 1+x x = y 1+x 1+y 1 + x = y +c 1+y 6. Solve 1 x sin 1 x + y = 0 = y integrating y sin 1 x 1 x = du u logy = - logu+logc yu = c ysin 1 x = c y = sin 1 x 1 x ; let u = sin 1 x,du = 7. Solve y-x+3x y e x 3 = 0 y x y +3x e x 3 = 0 d( x y ) + 3x e x 3 = x

8 integrating d( x y ) + 3x e x 3 = 0 x y + ex 3 = c Homogeneous Differential Equations A differential equation of the form = f(x,y) OR = g(x,y) f(y ) is called a Homogeneous x equation.where f(x,y) and g(x,y) are homogeneous equation of the same degree. Ex: = xy x +y In such case, y = vx = v+xdv 1. Solve = y + tan y x x Put y = vx = v+xdv v+x dv = v+tanv x dv = tanv dv tanv = x integrating dv logsinv = logx+logc sinv = xc, put v= y x sin y x = xc. Solve tanv = x +y = y x x Put y = vx = v+xdv v+x dv +v = v x dv = v -v dv v v dv v(v ) - 1 dv v +1 = x = x dv v = x 1 { = A + B v(v ) v - 1 logv +1 log (v ) = logx+logc v, A = - 1 B = 1

9 v v = x c, put v = y x v- = vx c y - = y x x x c y x = yx c 3. On putting y = vx, the homogenous differentioal equation x + y x + y = 0 Becomes xdv+(v+v ) = 0 4. Solve x-y = x + y Ans: xdv 1 + v = 0 5. Solve. = y 3 +3x y x 3 +3x y Ans: x = (1+3v ) v(v 1) dv Linear differential equations The linear equqtion in y of the first order is of the form are functions in x The general solution is y(i.f) = Where Integral Factor(I.F) = e P Q I. F + c The linear equqtion in x of the first order is of the form are functions in y The general solution is yxi.f) = Where Integral Factor(I.F) = e P Note : Integral factor of D.E is (I.F) Then, P = (I.F) I.F 1. Find the integral factor of (1- x ) (1- x ) P = x (1 x ) I.F = e P = e + x (1 x ), Q = x 1 x (1 x ) x Q I. F + c y = x 1 x (1 x ) (1 x ) = e log (1 x ) =. Find I.F of +ycotx = sinx I.F = e P = e cot = e log sinx = sinx 3. Find I.F of (1+x ) + y = etan 1 x +xy = x 1 x 1 (1 x ) +Px = Q(x), where P, Q +Py = Q(y), where P, Q

10 (1-+ x ) + 1 (1+x ) y = e tan 1x (1+ x ) I.F = e 1 (1+x ) = e tan 1 x 4. Find the solution of x + x (1+x ) y = 0 I.F = e (1+x ) = e log (1+x ) = (1+x ) The solution is, y(i.f) = Q I. F + c y(1+x ) = 0 + c y(1+x ) = c 5. If the integral factor of D.E is secx, find P(x) P(x) = (I.F) I.F = secxtanx secx = tanx 6. If e x is an integral factor of D.E. Find P(x) P(x) = (I.F) = 1 I.F e x 7. If e f x is IF of D.E, then P(x) = f (x) Exercise = e x 1. Solve, +y = x, Ans: ex y x + 1 = c. Find IF of +xy = x, ans: ex 3. Find IF of + y = x sin (x ) Ans : x 4. Solve +ytanx = sinx, Ans: y = cosx+ccos x 5. Find I.F of 3y = 6, Ans: e 3x 6. Find the solution of + y 3 =0, Ans: yex = Bernoulli s equation An equation of the form y n y n Put z = y 1 n dz = (1 n)y n 1 dz =y n (1 n) 3e x + c + py = Qyn, Where P,Q are functions in x only. + Py1 n = Q (1)

11 (1) 1 dz (1 n) 1. Solve, (x+1) +Pz = Q,which is linear equation in z +1 = e y e y e y x + 1 dz + ey = (1) Let z = e y dz = ey (1) (x+1) dz +z = (x+1) dz I. F = e + 1 (x+1) =(x+1) (x+1) z = (x+1) The general solution is, ze p = Qe P + c Z(x+1) = (x+1) Z(x+1) = x +c (x+1) +c. Find the I.F of (x y 3 + xy) = 1 Exercise x y 3 + xy = x 1 put z = - 1 x x dz = 1 x (1) dz +zy =y3 -y x = y (1) () I. F = e p = e y =e y 1. Find I F of + xtany = x3,ans: e x. Find the I.F of + ycosx = yn sinx, Ans: I F = e Equation of the first order with higher degree We shall denote by p 1 n sinx The D.E of the first order and nth degree, p n + A 1 p n 1 + A p n A A n = 0

12 Where A 1, A, A 3,.. A n denote functions of x and y In this case resolve the given equation into factor of first degree, (p-r 1 ) (p-r ) (p-r 3 ) (p-r n ) = 0 This implies, p =R 1,p =R,p =R 3, p=r n Solve the equation,then we get, φ 1 x, y, c 1 = 0, φ x, y, c = 0, φ 3 x, y, c 3 = 0,,φ n x, y, c n = 0, Where c 1, c, c 3,.., c n are constant The solution of D.E is, φ 1 x, y, c 1 φ x, y, c φ 3 x, y, c 3,φ n x, y, c n = 0 1. Solve, p 5p + 6 = 0 (Or) ( ) = 0 (p-)(p-3)= 0 p =, p = 3 =, = 3 =, = 3 Integrating, y = x+c 1,y = 3x+c y - x- c 1 = 0,y - 3x c = 0 The solution is,( y - x- c 1 )( y - 3x c ) = 0. Solve, p(p-y) = x(x+y) p py x xy = 0 p x y x + p = 0 (p+x)(p-x) y(x+p)=0 (p+x)(p-x-y) = 0 P = -x, p = x+y = -x y = -x + c 1 (y+x c 1 ) = 0 = x+y I. F = e =e x -y= x The solution, ye x = x e x + c ye x = xe x e x + c (y+x+1 - c e x ) = 0 the solution is, (y+x c 1 )(y+x+1 - c e x ) = 0

13 3. Solve, x p + 3xyp + y = 0 (TRB) x p + xyp + 3xyp + y = 0 xp xp + y + y(xp + y) = 0 px + y xp + y = 0 Px = -y, xp = -y Exercise x = -y logy = -logx+logc 1 (xy -c 1 ) = 0 xp = -y x = -y logy = -logx+logc 1 (yx -c ) = 0 The solution is, (xy -c 1 )(yx -c ) =0 4. Solve, p 7p + 10 = 0,Ans: ( y - x- c 1 )( y - 5x c ) = 0 5. Solve, p 7p + 1 = 0,Ans: ( y - 4x- c 1 )( y - 3x c ) = 0] NATIONAL ACADEM TRB MATHS COACHING CENTRE DHARMAPURI DIFFERENTIAL EQUATIONS TEST What is the order and degree of the differential equation d y ( (a) First order, second order (c) first order, first degree (b) second order,first degree ) =0 (d) second order, second degree. The differential equation derive from y = Ae x + Be x have the order,when A,B are constants (a) 3 (b) (c) 1 (d) None of these 3. The differential equation of y = Ae 3x + Be 5x is (a) y -8y+15y = 0 (b) y +8y+15y = 0 (c) y +8y = 0 (d) y = 0 4. The differential equation of x = Acos(pt-α) is (a) d x =0 (b) d x dt dt =-p x 5. The solution of = xy is (c) d x =0 dt (d) =-px dt (a) c= logy+x (b) c =logy-x (c) c =xy (d) c = ylogx 6. The degree of the differential equation of ( d y )3 + ( ) + sinx + 1 = 0 is (a) 1 (b) (c) 3 (d) 4 7. The number of arbitrary constants in the general solution of differential equation of fouth order is (a) 1 (b) (c) 3 (d) 4

14 8. A homogeneous differential equation of the form = f(x ) can be solve by making y the substitution (a) y = vx (b) v = yx (c) x = vy (d) x = v 9. Order and degree of differential equation are always (a) positive integer (b) Negative integer (c) integer (d) None of these 10. The differential equation of y = Acosx Bsinx is (a) y - y = 0 (b) y +y = 0 (c) y +y = 0 (d) y +x y = The differential equation of y = Ae x + Be x +3x is (a) y +x y = 0 (b) y + y = 3x (c) y - y = - 3x (d) y + y = x 1. Order and degree of differential equation ( d y )1 = (x )1 3 (a) (3,) (b) (1,3) (c) (,1) (d) (,3) 13. The solution of the differential equation (a) c = y+1 x 1 (b) c = y+1 x+1 = y+1 is x The general solution of differential equation =y x 3 (a) y = x 3 +k (b) y = x 4 +k (c) y = 4 x 3 +e x + k (c) c = (y+1)(x-1) (d) c(x-1) =y x 4 +k 15. The solution of the differential equation = ex y + x 3 e y is (a) x(x-y) + y(3-y) = c (b) 4e y = e x + x 4 + c (c) e x + e y = c (d) e x + e y + x 4 = c (d) y = 16. The general solution of differential equation sinxcosy cosxsiny = 0 (a) sinx = xsecy (b) xsecx = csecy (c) secx = cscey (d) secx = scey + c 17. The particular solution of y = xe x with initial condition y = 3 when x = 1 (a) x = ye x -e x +3 (b) y = xe x -e x +1 (c) y =xe x -e x +3 (d) y = e x (x+1) A curve passes through the point (0,0) with differential equation y = e 3x-y,then the equation of the curve is (a) 3e x = ye 3x +1 (b) e y = 3e 3x +1 (c) 3e y = e 3x (d) 3e y = e 3x The general solution of differential equation = ex (a) y = e x +c (b) y =e -x +c (c) y =-e x +c (d) x = logy+c 0. The differential eqution of the family parabolas y = 4ax is (a) = 4 ( ) (b) y = x (c) d y = 4 1. The differential eqution of the family of straight lines is (d) none of these (a) = 4 ( ) (b) y = x (c) d y =0 (d) None of these. Differential equation of the family of circles with center at origin & radius a is (a) X-y = 0 (b) y- x = 0 (c) x+y = 0 (d) y+x = 0

15 3. Differential equation of the family of circles which passes through the origin & whose centers are on the x-axis is (a) xy +x +y = 0 (b) xy +x - y = 0 (c) x +x - y = 0 (d) y +x - y = 0 4. The solution of ( 1+ x ) = 1 + y is (a) tan 1 y tan 1 x = tan 1 c (b) tan 1 y + tan 1 x = tan 1 c (c) tan 1 x tan 1 y = tan 1 c (d) tan 1( y x ) = tan 1 c 5. Solution of (xy +x)+(xy + y) = 0is (a) (x + 1) = c(y + 1) (b) (x 1) = c(y 1) (c) (x + 1)(y + 1) = c (d) (x 1)(y 1) = c -Material Available with Question papers- CONTACT ,