Particular Solutions Our eamples so far in this section have involved some constant of integration, K. We now move on to see particular solutions, where we know some boundar conditions and we substitute those into our general solution to give a particular solution..
Eample: 1 a. Find the particular solution for the differential equation given that (0) = 1. Solution: At first rearranging to separate variables: Now integrating i.e. ' d 6 6 d 6 d d (6 ) d d 6 d d
Eample: 1 (Contd.) For the integral involving, we use substitution method i.e. put u = 6 giving du = d. This means we'll replace d with ( 1/)du and integrate (1/u)du giving ln u, i.e. general solution is given b: 1 ln6 K Now, initial condition is when = 0, = 1; so we have:
Eample: 1 (Contd.) So, on substituting this back into equation of general solution ields: Particularsolution
Eample: 1 (Contd.) Checking our solution: substituting the particular solution back into the given differential equation : d d d d 6 6 4e (3 e and so ) Also, when = 0, = 3 e 0 = 1. So the particular solution is given b: 3 e
Eercise: 1 a. Find the particular solution for the differential equation given that () = e. Solution: At first rearranging to separate variables: Now integrating i.e. d d ln ln d d d d ln ln d d For the part, let u = ln, then du = d/. Hence, General solution is: ln(ln ) ln K
Eercise: 1 (contd.) Substituting = when = e into general solution's equation gives: So the particular solution is given b:
Solution of differential equations 3. Homogeneous equations If an equation in the differential form M (, ) d N(, ) d 0 has the propert that n n M ( t, t) t M (, ) and N( t, t) t N(, ) We sa it has homogenous coefficients or it is a homogenous equation. It should be noted that homogenous differential equation can alwas be reduced to a separable equation through a simple algebraic substitution.
Solution of differential equations 3. Homogeneous equations Before pursuing a method of solution for homogenous differential equation,, lets eamine the nature of homogenous function If f(t,t) function f of degree n. t n (,) for some real number n, then the (,) is said to be a homogenous function Eample 1: f f (, ) 3 5 ( t, t) t t t 3 tt 5t 3 t 5t 3 5 tf (, ) Thus the function is homogenous of degree 1
Eamples: Homogeneous function Eample : ), ( ), ( ), ( 3/ 3 3 3/ 3 3 3 3 3 3 3 3 f t t t t t t t t f f Thus the function is homogenous of degree 3/ Eample 3: Since, ), ( ), ( 1 ), ( 1 ), ( t t t f t f t t t t t f f Thus the function is not homogenous
Eamples: Homogeneous function (Contd.) Eample 4: f (, ) f ( t, t) t t 4 4 4 t 0 f (, ) Thus the function is homogenous of degree 0 It can be seen from Eample 3 and 4 that, a constant added to function destros homogeneit, unless the function is homogenous of degree 0.
Eamples: Homogeneous function (Contd.) A homogenous function can also be recognized b eamining the total degree of each term. Degree 1 Degree Eample 1: 3 f (, ) 6 Degree Degree 3 Thus the function is homogenous of degree 4 Eample : f (, ) Degree Degree 1 Thus the function is not homogenous, since the degrees of two terms are different.
Homogeneous Function A function f (, ) in and is called a homogenous function, if the degrees of each term are equal. Eamples: g, = - + 3 f, = +3 + is a homogenous function of degree is a homogenous function of degree 3
Homogenous Differential Equations d = ƒ, d g, where f (, ) and g(, ) is a homogenous functions of the same degree in and, then it is called homogenous differential equation. Eample: 3 d +3 = d 3 is a homogenous differential equation as 3 +3 and 3 both are homogenous functions of degree 3.
3. Homogeneous equations: Method of Solution An equation of the form M (, ) d N(, ) d 0 Where, M and N have same degree of homgenit, can be reduced to separable variables b the substitution = u, where u is new dependent variable. Eample 1: Solve d d 0 It can be seen that both M (, ) and N (, ) are homogenous of degree. If we let = u, then
Solve Eample 1 (Contd.) d d 0 let = u u d u ud du d ( (1 (1 (1 u u u ud u) d ) d ) d du : u ud u u ( u 3 Differentiating using u (1 u ) d ) d u) du 3 (1 3 (1 0 u) du du u) du product rule 0 0 Now rewriting above equation in separable form 1 1 u u du d 0
Now separable form Eample: 1 (Contd.) 1 u d equation is du 0 1 u Rearranging termsinu d 1 0 1 du u Now integration will ield general solution, i.e. 1 1 du u u ln1 u ln1 ln ln d K K 0 0 0 ( u u /)
Working Rule to solve a Homogenous DE: 1. Put the given equation in the form M (, ) d N(, ) d 0 (1). Check M and N are Homogeneous function of the same degree. 3. Let u.
4. Differentiate = u to get d du u d d 5. Put this value of d/d into (1) and solve the equation for u b separating the variables. 6. Replace u b / and simplif.
Eample. Solve the DE (3 ) d d 0 It isa homogenousequationof degree. That is d d (3 ) Let = u. Hence we get u du d (3 u u ) u (3 u )
Eample: (Contd.) or du d u (3 u ) u u (3 u u 3 ) Separating the variables, we get (3 u 3 u ) u Integrating we get du d ) (3 u du u 3 u d
We epress the LHS integral b partial fractions. We get Eample: (Contd.) 3 u u 1 1 u 1 1 du d ln c 3 or u ( u 1)( u 1) c, c an arbitrar constant. Noting u = /, the solution is: ( )( ) c 3, c an arbitrar constant or c, 3 c an arbitrar constant
Eercises d a. Solve Ans : ln K d b d d. Solve Ans : ln1 4ln K c. Solve ' 3 Ans : ln 1 ln K d. Solve d ( ) d 0 Ans : ln ln K
d Solve d e Eample: 4 / The above equation can be rewritten as: If we let = u, then d = du+ ud subject to (1) d d ud du u u/ du e u u d d Now implementing variable seperation e u du Now at = 1, = 1 1 0 1 e ln1 d e K 1. 1 u General Solution K K e 1 ln e / ln e u e / e Particular Solution 1 0
Eercise: 1 d 3 3 Solve subject to (1). d The above equation can be rewritten as: If we let = u, then d = du+ ud General Solution d d 3 3 Particular Solution
Partial Derivatives Let f(,) be a function with two variables. If we keep constant and differentiate f (assuming f is differentiable) with respect to the variable, we obtain what is called the partial derivative of f with respect to which is denoted b: f or Similarl If we keep constant and differentiate f (assuming f is differentiable) with respect to the variable, we obtain what is called the partial derivative of f with respect to which is denoted b f or f f
Partial Derivatives: Eamples E 1. E. f (, ) 3 ln f 6 ln f 1 3 g(, ) e g 1 e
Partial Derivatives: Eamples E 3. f (,, z) 4 z 3 f 4 3 z f 3 3 4 z f z 3 z 4
Eample 1: Find the partial derivatives f and f if f(, ) is given b f(, ) = + +
Eample : Find f and f if f(, ) is given b f(, ) = sin( ) + cos
Eample 3: Find f and f if f(, ) is given b f(, ) = e
Eample 4: Find f and f if f(, ) is given b f(, ) = ln ( + )
Second-Order Partial Derivatives (f, f ) E 1. 3 5 f (, ) ln f 0 f 6 3 3 f f 1 6
Notation for Partial Derivatives f f f f means means means means f f f f
Total Derivative The total derivative (full derivative) of a function, f, of several variables, e.g., t,,, etc., with respect to one of its input variables, e.g., t, is different from the partial derivative. Calculation of the total derivative of f with respect to t does not assume that the other arguments are constant while t varies; instead, it allows the other arguments (i.e. & ) to depend on t. The total derivative adds in these indirect dependencies to find the overall dependenc of f on t. For eample, the total derivative of f(t,,) with respect to t is Consider multipling both sides of the equation b the differential dt
4. EXACT DIFFERENTIAL EQUATIONS A first order DE M (, ) d N(, ) d 0 is called an eact DE if there eits a function f(, ) such that left side is an eact differential, i.e. df M (, ) d N(, ) d Here df is the total differential of f(, ) and equals f f d d
4. EXACT DE (Contd.) Hence the given DE becomes df = 0 Integrating, we get the solution as f(, ) = c, c an arbitrar constant Eample: The DE d d 0 is eact as it is d () = 0 Hence the solution is: = c
Eample The DE d d 0 is eact as it is d ( + ) = 0 Hence the solution is: + = c 1 sin( ) d sin( ) d 0 Eample 3 The DE is eact as it is d(cos( )) 0 Hence the solution is: cos( ) c
Test for eactness Suppose M (, ) d N(, ) d 0 is eact. Hence there eists a function f(, ) such that f f M (, ) d N(, ) d df d d f f Hence M, N Assuming all the nd order mied derivatives of f(, ) are continuous, we get
M f f N Thus a necessar condition for eactness is M N
We saw a necessar condition for eactness is M N We now show that the above condition is also sufficient for M d + N d = 0 to be eact. That is, if, f f M N then there eists a function f(, ) such that M N
Integrating f M partiall w.r.t., we get f (, ) M d g( ) where g() is a function of alone We know that for this f(, ),. (*) f Differentiating (*) partiall w.r.t., we get f M d g( ) = N gives N
or g( ) N M d (**) We now show that the R.H.S. of (**) is independent of and thus g() (and so f(, )) can be found b integrating (**) w.r.t.. N M d N M d N M d N M = 0
Note (1) The solution of the eact DE d f = 0 is f(, ) = c. Note () When the given DE is eact, the solution f(, ) = c is found as we did in the previous theorem. That is, we integrate M partiall w.r.t. to get f (, ) M d g( ) We now differentiate this partiall w.r.t. and equating to N, find g () and hence g(). The following eamples will help ou in understanding this.
Eample: Test whether the following DE is eact. If eact, solve it. 1 d (3 7) d 0 It can be seen that comparing with standard differential form of M, d N(, ) d 0 M 1 & N (3 7) Now checking whether given equation is eact or not M 0 & N 0 M i.e. N Hence, the given equation is eact, and there eist a function f(,), such that: f M & f N
f Step : Now partiall integrating keeping constant while integrating Eample (Contd.) with M respect to to obtain f(,) i.e. f (, ) Md ( 1) d g( ) Eq :1 Here; g() is function of alone, and represents constant of integration Step 3: Now we know for this function f(,), partiall differentiating f(,) w.r.t. f f Now we alsoknow g() 3 7 g( ) g' ( ) N g'() 37; Now integrating both sidesto obtain g() f N Therefore, (3 7), Thereb equating the two terms ields
Step 4: Now substituting this value of g() into equation 1 ields general solution of the sstem Eample (Contd.) equations eactdifferentail for c f c f ), ( 7 3 ), (
Eample 1: Test whether the following DE is eact. If eact, solve it. ( ) d d 0 Here M, N ( ) M 1 N Hence eact. Now f (, ) M d d g( )
Eample: 1 (contd.) Differentiating partiall w.r.t., we get Hence f g( ) N g( ) Integrating, we get g( ) ln (Note that we have NOT put the arb constant ) Hence f (, ) g( ) ln Thus the solution of the given DE is f (, ) c or ln c, c an arb const.
Eample : Test whether the following DE is eact. If eact, solve it. 4 3 ( sin ) d (4 cos ) d 0 Here 4 3 M sin, N 4 cos M 3 8 cos N Hence eact. Now 4 f (, ) M d ( sin ) d 4 sin g( )
Eample: (contd.) Differentiating partiall w.r.t., we get f 3 4 cos g( ) Integrating, we get g() = c Hence g( ) 0 (Note that we have NOT put the arb constant ) Hence f (, ) 4 sin g( ) 4 sin Thus the solution of the given d.e. is f (, ) c 3 N 4 cos or 4 sin c, c an arb const.
In the previous problems, we found f(, ) b integrating M partiall w.r.t. and then f equated to N. We can reverse the roles of and. That is we can find f(, ) b integrating N partiall w.r.t. and then equate f to M. The following problem illustrates this.
Eample 3: Test whether the following DE is eact. If eact, solve it. (1 sin ) d cos d 0 Here M M sin ; 1 sin, N Hence eact. Now f (, ) N d cos d N cos 4 cos sin sin cos g( )
Eample: 3 (contd.) Differentiating partiall w.r.t., we get f cos sin g( ) M 1 sin Integrating, we get g( ) gives g( ) 1 (Note that we have NOT put the arb constant ) Hence f (, ) cos g( ) cos Thus the solution of the given d.e. is f (, ) c or cos c, c an arb const. sin g( )
Eercise: Solve (e cos )d+(e cos + )d = 0. Solution: This DE is eact because M/ = e + sin cos = N/ Hence a function f eists, and f/ = e cos + that is, f (, ) f e d cosd e e sin cos g( ) g' ( ) e d cos
Thus g() = 0, Eercise: (contd.) Now integration ields g() = c. The general solution is e sin + + c = 0
d d Solve Solution: Eercise: 3 cos sin, (0) (1 ) Rewrite the DE in the form (cos sin ) d + (1 ) d = 0 Since M/ = = N/ (This DE is eact) Now f/ = (1 ) f(, ) = ½ (1 ) + g() f/ = + h() = cos sin
Eercise: 3 (contd.) We have g() = cos sin => Hint: Let u=cos => du = - sind g' ( ) g() = -½ cos Thus ½ (1 ) ½ cos = c 1 or (1 ) cos = c (cos)( where c = c 1. Now (0) =, so c = 3. The solution is (1 ) cos = 3 sin d)
Eercises Test whether the following DE is eact. If eact, solve it. 1 ( cos ) ' 0 a. 4 Ans : sin c d b. d (3 3 ) d 0 d d 0 (3 ) d Ans : 3 c
Eercises Test whether the following DE is eact. If eact, solve it. 3 d ( 4) d 0 c. Ans : 3 4 c d. 3 sin d 3 cosd 0 Ans : 3 cos c
d e. e 6 d Eercises Test whether the following DE is eact. If eact, solve it. Ans : e e 3 c f d d 0. Ans:Not eact; though it is homogenous of degree
Eercises: Initial value problem Test whether the following DE is eact. If eact, solve it. g. cos sin - d (1 ) d 0; (0) Ans : (1 ) cos 3 h. d 1d 0; (1) 1 Ans: 3 1 3 4 3
5. Linear Equations A linear first order equation is an equation that can be epressed in the form d a1( ) a0( ) b( ), (1) d where a 1 (), a 0 (), and b() depend onl on the independent variable, not on.
Linear Equations (contd.) We assume that the function a 1 (), a 0 (), and b() are continuous on an interval and that a 1 () 0 on that interval. Then, on dividing b a 1 (), we can rewrite equation (1) in the standard form d d P( ) Q( ) () where P(), Q() are continuous functions on the interval.
Let s epress equation () in the differential form P( ) Q( ) d d 0 (3) If we test this equation for eactness, we find M N P( ) and 0 Consequentl, equation(3) is eact onl when P() = 0. It turns out that an integrating factor, which depends onl on, can easil be obtained the general solution of (3).
Multipl (3) b a function () and tr to determine () so that the resulting equation ( ) P( ) ( ) Q( ) d ( ) d 0 (4) is eact. M N d ( ) P( ) and ( ) d We see that (4) is eact if satisfies the DE d ( ) P( ) ( ) (5) d ( ) ep P( ) d (6) Which is our desired IF
In (), we multipl b () defined in (6) to obtain d ( ) P( ) ( ) ( ) Q( ) (7) d We know from (5) P( ) ( ) d d and so (7) can be written in the form d d d d ( ) ( ) ( ) Q( ) d d ( ) ( ) Q( ) (8)
Integrating (8) w.r.t. gives ( ) ( ) Q( ) d C and solving for ields ep ( ) ( ) ( ) P d Q d C
Working Rule to solve a LDE: 1. Write the equation in the standard form d d P( ) Q( ). Calculate the IF () b the formula ( ) ep P( ) d 3. Multipl the equation in standard form b () and recalling that the LHS is just d ( ), d obtain
d ( ) ( ) Q( ) d 4. Integrate the last equation and solve for b dividing b ().
Sol: d d Solve Eample: 1 4 Step 1. Write the equation in the standard form 4 d d 5 P( ) Q( ) e 6 e Dividing throughout b Now comparing the above equation with the standard form ields: P( ) 4
Eample: 1(contd.) Step. Calculate the IF () b the formula e P( ) d e 4 / d e 4ln e ln 4-4 ( using ln propert) ( b lnn N ) Eample. Find the natural logarithm i.e. ln of 9.178 Solution: ln (= 9.178) =.168 Now, checking e.168 =.7188188.168 = 9.178 i.e. e ln 9.178 = 9.178 or e ln =
Eample: 1(contd.) Step 3. Now multipling the equation obtained in step 1 b I.F. i.e. () = -4 d d 4 4 5 e Now it can seen that above equation is eact as: M N ( 4 ( 4 ) 5 4 e 5 ) 4 It can also be seen that left side of equation obtained in step 3, is the derivative of integrating factor and dependent variable, i.e. d 5 d d d 4 4 4 5
Eample: 1(contd.) Step 4. Equation obtained in step 3 can be rewritten as: d d 4 e Step 5. Now integrating both sides will ield: d 4 e d Need to use integration b parts on R.H.S. 4 e e c General solution will be: 5 e 4 e c 4
Solve Eercises d 3-3 Ans : ce Note () e a. 3 d 0 d 3 3 b. 7 3 Ans : c Note () d d c. d 9 0 7 c Ans : Note () 9 9-4 d d 0 d. Ans : k Note ()