Advanced Engineering Math. Problem set 2.7 : 4 ~ 6

Problem 4 ~ 6의 경우 Homogeneous Soltn. 구하는 연습이 가능하고, Nonhomogeneous Soltn. 구할 때 Modification rule 또한 적용해 볼 수 있는 문제이다. 그동안 공부한 것들을 충분히 복습하여 내 것으로 만들자.

 

Problem 4.

Find the solution

$$ y'' - 9 y = 18 cos \pi x $$

1) Solution of homogeneous ODE

 Let's solve the equation : $ y'' - 9 y = 0 $

 By the characteristic equation, there are two real roots, $ \lambda = 3 \ or \ -3 $

 Hence, the solution of homogeneous ODE

$$ \therefore y_{h} = c_{1} e^{-3x} + c_{2} e^{3x} $$

2) Solution of Nonhomogeneous ODE

 Let's consider that $ y_{p} = K cos \pi x + M sin \pi x $ by basic rule of the method of undetermined coef..

$$ y_{p} = K cos \pi x + M sin \pi x $$

$$ y'_{p} = - \pi K sin \pi x + \pi M cos \pi x $$

$$ y''_{p} = - \pi ^{2} K cos \pi x - \pi ^{2} M sin \pi x $$

 Substitute into original equation, and we will get $ M = 0 $ and $ K = - \frac{18}{9 + \pi ^{2}}

$$ \therefore c_{1} e^{-3x} + c_{2} e^{3x} - \frac{18}{9 + \pi ^{2}} cos \pi x $$

 

Problem 5.

Find the solution

$$ y'' + 4y' + 4y = e^{-x} cos x $$

1) Solution of homogeneous ODE

Let's solve the equation : $ y'' + 4y' + 4y = 0 $

By the characteristic equation, there is double roots, $ \lambda = -2 $

The first solution of homogeneous of ODE is $ y_{1} = c_{1} e^{-2x} $

It is need to be found 2nd solution of homogen. ODE,

 ※ 하나의 기저를 알고 있기 때문에 계수감소법(Reduction of order)를 이용하여 나머지 해를 구할 수 있다.

 ※ 복습겸 전개를 해보자.

Let's assume the solution that : $ y_{2} = u y_{1} $

$$ y_{2} = u y_{1} $$

$$ y'_{2} = u' y_{1} + u y'_{1} $$

$$ y''_{2} = u'' y_{1} + 2 u' y'_{1} + u y''_{1} $$

$$ u''y_{1} + 2 u'y'_{1} + uy''_{1} + 4 u'y_{1} + 4 uy_{1} + 4 u y_{1} = 0 $$

In short,

$$ u'' y_{1} + 2 u' y'_{1} + 4 u'y_{1} = 0 $$

because $ u(y''_{1} + 4 y'_{1} + 4 y_{1}) = 0 $

Integrate, 

$$ y_{2} = x e^{-2x} $$

Hence,

$$ y_{h} = c_{1} e^{-2x} + c_{2} x e^{-2x} $$

2) Solution of Nonhomogeneous ODE

Let's apply Basic Rule of method of undetermined coefficients.

Assume that :

$$ y_{p} = e^{-x} ( K cos x + M sin x) $$

$$ y'_{p} = - K e^{-x} ( cos x + sin x ) + M e^{-x} ( cos x - sinx ) $$

$$ y''_{p} = 2K e^{-x} sinx - 2 Me^{-x} cos x $$

※ 전개할 때 조심하자... 미리미리 정리하고 전개해야 안 헷갈린다.. 

Hence, the equation will be :

$$ 2 K e^{-x} sin x - 2 M e ^{-x} cos x + 4 ( -Ke^{-x}(cosx + sinx) + M e^{-x}(cosx - sinx)) + 4 e^{-x}(Kcosx + Msinx) = e^{-x} cos x $$

The parameter K, M will be :

$$ K = 0 , M = \frac{1}{2} $$

Therefore,

$$ y_{p} = e^{-x} (\frac{1}{2} sin x ) $$

Finally, we can get the solution : 

$$ \therefore y(x) = c_{1} e^{-2x} + c_{2} x e^{-2x} + \frac {1}{2} e^{-x} sin x $$

 

Problem 6.

Find the solution

$$ y'' + y' + (\pi^{2} + 1/4) y = e^{-x/2} sin \pi x $$

1) Solution of homogeneous ODE

$ \lambda = - \frac{1}{2} \pm \pi i $

$$ \therefore y_{h} = e^{-x/2} (A cos \pi x + B sin \pi x) $$

2) Solution of Nonhomogeneous ODE

$ y_{p} $ is expected to be $ e^{ax} ( K cos \pi x + M sin \pi x ) $, but is is already exist when checking $ y_{h} $

So, we need to apply the modification rule, Let's multiply $ x $ and solve it.

아래는.. 생략... 작성하는데 시간이 너무 오래 걸린다.