Question

Apply the method of undetermined coefficients to find a particular solution to the following system.wing system.

y" - y' + y = sin x.
y" - 3y' + 2y = e^x sin x.
y" + y = x cos(2x).

Answer 1

• 
  `y''-y'+y=\sin x`

The corresponding homogeneous ODE has characteristic equation
`r^2-r+1=0` with roots at
`r=\frac{1\pm\sqrt3}2`, thus admitting the characteristic solution

`y_c=C_1e^x\cos\frac{\sqrt3}2x+C_2e^x\sin\frac{\sqrt3}2x`

For the particular solution, assume one of the form

`y_p=a\sin x+b\cos x`

`{y_p}'=a\cos x-b\sin x`

`{y_p}''=-a\sin x-b\cos x`

Substituting into the ODE gives

`(-a\sin x-b\cos x)-(a\cos x-b\sin x)+(a\sin x+b\cos x)=\sin x`

`-b\cos x+a\sin x=\sin x`

`\implies a=1,b=0`

Then the general solution to this ODE is

`\boxed{y(x)=C_1e^x\cos\frac{\sqrt3}2x+C_2e^x\sin\frac{\sqrt3}2x+\sin x}`

• 
  `y''-3y'+2y=e^x\sin x`

`\implies r^2-3r+2=(r-1)(r-2)=0\implies r=1,r=2`

`\implies y_c=C_1e^x+C_2e^(2x)`

Assume a solution of the form

`y_p=e^x(a\sin x+b\cos x)`

`{y_p}'=e^x((a+b)\cos x+(a-b)\sin x)`

`{y_p}''=2e^x(a\cos x-b\sin x)`

Substituting into the ODE gives

`2e^x(a\cos x-b\sin x)-3e^x((a+b)\cos x+(a-b)\sin x)+2e^x(a\sin x+b\cos x)=e^x\sin x`

`-e^x((a+b)\cos x+(a-b)\sin x)=e^x\sin x`

`\implies\begin{cases}-a-b=0\\-a+b=1\end{cases}\implies a=-\frac12,b=\frac12`

so the solution is

`\boxed{y(x)=C_1e^x+C_2e^(2x)-\frac{e^x}2(\sin x-\cos x)}`

• 
  `y''+y=x\cos(2x)`

`r^2+1=0\implies r=\pm i`

`\implies y_c=C_1\cos x+C_2\sin x`

Assume a solution of the form

`y_p=(ax+b)\cos(2x)+(cx+d)\sin(2x)`

`{y_p}''=-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x)`

Substituting into the ODE gives

`(-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x))+((ax+b)\cos(2x)+(cx+d)\sin(2x))=x\cos(2x)`

`-(3ax+3b-4c)\cos(2x)-(3cx+3d+4a)\sin(2x)=x\cos(2x)`

`\implies\begin{cases}-3a=1\\-3b+4c=0\\-3c=0\\-4a-3d=0\end{cases}\implies a=-\frac13,b=c=0,d=\frac49`

so the solution is

`\boxed{y(x)=C_1\cos x+C_2\sin x-\frac13x\cos(2x)+\frac49\sin(2x)}`