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

$$y^{″}-9y=18cos\pi x$$

1) Solution of homogeneous ODE

 Let's solve the equation : $y^{″}-9y=0$

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

 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 $y_p=Kcos\pi x+Msin\pi x$ by basic rule of the method of undetermined coef..

$$y_p=Kcos\pi x+Msin\pi x$$

$$y_p^{\prime }=-\pi Ksin\pi x+\pi Mcos\pi x$$

$$y_p^{″}=-{\pi }^{2}Kcos\pi x-{\pi }^{2}Msin\pi x$$

 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

$$y^{″}+4y^{\prime }+4y=e^{-x}cosx$$

1) Solution of homogeneous ODE

Let's solve the equation : $y^{″}+4y^{\prime }+4y=0$

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 : $y_2=u{y}_1$

$$y_2=u{y}_1$$

$$y_2^{\prime }=u^{\prime }{y}_1+u{y}_1^{\prime }$$

$$y_2^{″}=u^{″}{y}_1+2u^{\prime }{y}_1^{\prime }+u{y}_1^{″}$$

$$u^{″}{y}_1+2u^{\prime }{y}_1^{\prime }+u{y}_1^{″}+4u^{\prime }{y}_1+4u{y}_1+4u{y}_1=0$$

In short,

$$u^{″}{y}_1+2u^{\prime }{y}_1^{\prime }+4u^{\prime }{y}_1=0$$

because $u(y_1^{″}+4y_1^{\prime }+4y_1)=0$

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 :

$$y_p=e^{-x}(Kcosx+Msinx)$$

$$y_p^{\prime }=-K{e}^{-x}(cosx+sinx)+M{e}^{-x}(cosx-sinx)$$

$$y_p^{″}=2K{e}^{-x}sinx-2M{e}^{-x}cosx$$

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

Hence, the equation will be :

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

The parameter K, M will be :

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

Therefore,

$$y_p=e^{-x}(\frac{1}{2}sinx)$$

Finally, we can get the solution : 

$$\therefore y(x)=c_1{e}^{-2x}+c_{2}x{e}^{-2x}+\frac{1}{2}{e}^{-x}sinx$$

 

Problem 6.

Find the solution

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

1) Solution of homogeneous ODE

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

$$\therefore y_h=e^{-x/2}(Acos\pi x+Bsin\pi x)$$

2) Solution of Nonhomogeneous ODE

$y_p$ is expected to be $e^{ax}(Kcos\pi x+Msin\pi x)$, 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.