Non-homogeneous equations (Sect. 3.6).

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1 Non-homogeneous equations (Sect. 3.6). We study: y + p(t) y + q(t) y = f (t). Method of variation of parameters. Using the method in an example. The proof of the variation of parameter method. Using the method in another example. Method of variation of parameters. Remarks: This is a general method to find solutions to equations having variable coefficients and non-homogeneous with a continuous but otherwise arbitrary source function, y + p(t) y + q(t) y = f (t). The variation of parameter method can be applied to more general equations than the undetermined coefficients method. The variation of parameter method usually takes more time to implement than the simpler method of undetermined coefficients.

2 Method of variation of parameters. Theorem (Variation of parameters) Let p, q, f : (t, t ) R be continuous functions, let y, y : (t, t ) R be linearly independent solutions to the homogeneous equation y + p(t) y + q(t) y = 0, and let W y y be the Wronskian of y and y. If the functions u and u are defined by u (t) = y (t)f (t) W y y (t) dt, u y (t)f (t) (t) = W y y (t) dt, then the function y p = u y + u y is a particular solution to the non-homogeneous equation y + p(t) y + q(t) y = f (t). Non-homogeneous equations (Sect. 3.6). We study: y + p(t) y + q(t) y = f (t). Method of variation of parameters. Using the method in an example. The proof of the variation of parameter method. Using the method in another example.

3 Using the method in an example. Find the general solution of the inhomogeneous equation y 5y + 6y = e t. Solution: First: Find fundamental solutions to the homogeneous equation. The characteristic equation is { r 5r + 6 = 0 r = ( ) r = 3, 5 ± 5 4 r =. Hence, y (t) = e 3t and y (t) = e t. Compute their Wronskian, W y y (t) = (e 3t )(e t ) (3e 3t )(e t ) W y y (t) = e 5t. Second: We compute the functions u and u. By definition, u = y f W y y, u = y f W y y. Using the method in an example. Find the general solution of the inhomogeneous equation y 5y + 6y = e t. Solution: Recall: y (t) = e 3t, y (t) = e t, W y y (t) = e 5t, and u = y f W y y, u = y f W y y. u = e t (e t )( e 5t ) u = e t u = e t, u = e 3t (e t )( e 5t ) u = e t u = e t. Third: The particular solution is y p = ( e t )(e 3t ) + (e t )(e t ) y p = e t. The general solution is y(t) = c e 3t + c e t + e t, c, c R.

4 Non-homogeneous equations (Sect. 3.6). We study: y + p(t) y + q(t) y = f (t). Method of variation of parameters. Using the method in an example. The proof of the variation of parameter method. Using the method in another example. The proof of the variation of parameter method. Proof: Denote L(y) = y + p(t) y + q(t) y. We need to find y p solution of L(y p ) = f. We know y and y solutions of L(y ) = 0 and L(y ) = 0. Idea: The reduction of order method: Find y proposing y = uy. First idea: Propose that y p is given by y p = u y + u y. We hope that the equation for u and u will be simpler than the original equation for y p, since y and y are solutions to the homogeneous equation. Compute: y p = u y + u y + u y + u y, y p = y + u y + u y + y + u y + u y.

5 The proof of the variation of parameter method. Proof: Then L(y p ) = f is given by [ u y + u y + u y + y + u y + u y ] p(t) [ u y + u y + u y + u y ] + q(t) [ u y + u y ] = f (t). y + y + (u y + u y ) + p (u y + u y ) +u (y + p y + q y ) + u (y + p y + q y ) = f Recall: y + p y + q y = 0 and y + p y + q y = 0. Hence, y + y + (u y + u y ) + p (u y + u y ) = f Second idea: Look for u and u that satisfy the extra equation u y + u y = 0. The proof of the variation of parameter method. Proof: Recall: u y + u y = 0 and y + y + (u y + u y ) + p (u y + u y ) = f. These two equations imply that L(y p ) = f is y + y + (u y + u y ) = f. From u y + u y = 0 we get [ u y + u y ] = 0, that is y + y + (u y + u y ) = 0. This information in L(y p ) = f implies u y + u y = f. Summary: If u and u satisfy u y + u y = 0 and u y + u y = f, then y p = u y + u y satisfies L(y p ) = f.

6 The proof of the variation of parameter method. Proof: Summary: If u and u satisfy u y + u y = 0 and u y + u y = f, then y p = u y + u y satisfies L(y p ) = f. The equations above are simple to solve for u and u, u = y u u y y y y u = f u (y y y y ) = f. y y Since W y y = y y y y, u = y f W y y u = y f W y y. Integrating in the variable t we obtain u (t) = This establishes the Theorem. y (t)f (t) W y y (t) dt, u (t) = y (t)f (t) W y y (t) dt, Non-homogeneous equations (Sect. 3.6). We study: y + p(t) y + q(t) y = f (t). Method of variation of parameters. Using the method in an example. The proof of the variation of parameter method. Using the method in another example.

7 Using the method in another example. Find a particular solution to the differential equation t y y = 3t, knowing that the functions y = t and y = /t are solutions to the homogeneous equation t y y = 0. Solution: First, write the equation in the form of the Theorem. That is, divide the whole equation by t, y t y = 3 t f (t) = 3 t. We know that y = t and y = /t. Their Wronskian is ( ) ( ) W y y (t) = (t ) t (t) t W y y (t) = 3. Using the method in another example. Find a particular solution to the differential equation t y y = 3t, knowing that the functions y = t and y = /t are solutions to the homogeneous equation t y y = 0. Solution: y = t, y = /t, f (t) = 3 t, W y y (t) = 3. We now compute y and u, u = ( 3 ) t t 3 = t 3 t 3 u = ln(t) + 6 t, ( u = (t ) 3 ) t 3 = t + 3 u = 3 t3 + 3 t.

8 Using the method in another example. Find a particular solution to the differential equation t y y = 3t, knowing that the functions y = t and y = /t are solutions to the homogeneous equation t y y = 0. Solution: The particular solution ỹ p = u y + u y is [ ỹ p = ln(t) + 6 t ] (t ) + 3 ( t3 + t)(t ) ỹ p = t ln(t) t + 3 = t ln(t) + 3 t ỹ p = t ln(t) + 3 y (t). A simpler expression is y p = t ln(t) +. Using the method in another example. Find a particular solution to the differential equation t y y = 3t, knowing that the functions y = t and y = /t are solutions to the homogeneous equation t y y = 0. Solution: If we do not remember the formulas for u, u, we can always solve the system u y + u y = 0 t u + u u y + u y = f. t = 0, t u + u ( ) t = 3 t. u = t 3 u t u + t u = 3 t u = t 3t 3 u = t + 3.