Answer to Question #122593 in Differential Equations for jse

Question

1. Solve the given differential equation by undetermined coefficients.
y'' − 2y' + 5y = e^x sin x

y(x) =____

2. Solve the given differential equation by undetermined coefficients.
y'' − y' − 2y = e^2x

y(x) =___

Answer 1

1.The given equation is $(D^{2}-2D+5)y = e^{x}\sin x$.

The auxiliary equation is $m^{2}-2m+5=0.$ Solving for m, we get $m=1\pm 2i$.

Thus, the complementary function is $C.F = e^{x}(c_{1}\cos 2x + c_{2}\sin 2x).$

To find the particular integral, let $y_{p} = Ae^{x}\sin x$ be the trial solution.

Then,

$y'_{p} = Ae^{x}(\cos x + \sin x)~\text{and}~\\ y''_{p} = A\bigg(e^{x}(-\sin x+\cos x )+e^{x}(\cos x + \sin x)\bigg)\\ ~~~~~= 2Ae^{x}\cos x$

Therefore,

$2Ae^{x}\cos x - 2\bigg(Ae^{x}(\cos x+\sin x)\bigg)+5Ae^{x}\sin x = e^{x}\sin x\\ A = \dfrac{1}{3}$

Hence the particular integral is $y_{p} = \dfrac{1}{3}e^{x}\sin x$ .

The general solution is $y=e^{x}(c_{1}\cos 2x+c_{2}\sin 2x)+\dfrac{1}{3}e^{x}\sin x$

2.The given equation is $(D^{2}-D-2)y=e^{2x}.$

The auxiliary equation is $m^{2}-m-2=0$. Solving for m, we get $m=-1,2$.

Thus, the complementary function is $C.F=c_{1}e^{-x}+c_{2}e^{2x}$.

To find the particular integral, since $g(x) =e^{2x}$ is a linear combination of a term in complementary function we choose $y_{p} = Axe^{2x}$ as the trial particular solution.

Thus,

$y'_{p}=A(2xe^{x}+e^{2x})=Ae^{2x}(2x+1)\\ y''_{p}=A\bigg(2(2xe^{x}+e^{2x})+2e^{2x}\bigg)=4Ae^{2x}(x+1)\\$

Therefore,

$4Ae^{2x}(x+1)-Ae^{2x}(2x+1)-2Axe^{2x}=e^{2x}\\ A(4x+4-2x-1-2x)=1\\ A=\dfrac{1}{3}$

Hence the particular integral is, $y_{p}=\dfrac{1}{3}xe^{2x}$

The general solution is, $y=c_{1}e^{-x}+e^{2x}(\frac{x}{3}+c_{2})$