-17.2.9 Solve \ y''-y&=2e^{-x}+3e^{2} by undetermined coefficients

[karush]

ok going to see if i can do the steps
$\textrm{17.2.9 Solve the given equation by the method of undetermined coefficients.}\\$
\begin{align*}\displaystyle
y''-y&=2e^{-x}+3e^{2}
\end{align*}
$\textit{ textbook answer is:}\\$
\begin{align*}\displaystyle
y&=c_1e^{-x}+c_2 e^{-x}-xe^{-x}-3x^{2}-6\\
\end{align*}

 

[MarkFL]

Could if be that the given ODE is:

$$y''-y=2e^{-x}+3x^2$$

And the textbook answer is equivalent to:

$$y(x)=c_1e^x+c_2e^{-x}-xe^{-x}-3x^2-6$$ ?

 

[MarkFL]

I will proceed as if what I posted is correct, and give my solution below:

Spoiler

We see that the characteristic roots are:

$$r=\pm1$$

And so the homogeneous solution is:

$$y_h(x)=c_1e^{x}+c_2e^{-x}$$

We will assume our particular solution then will take the form:

$$y_p(x)=Axe^{-x}+Bx^2+Cx+D$$

Differentiating, we find:

$$y_p''(x)=-2Ae^{-x}+Axe^{-x}+2B$$

Substitution into the ODE gives us:

$$-2Ae^{-x}+Axe^{-x}+2B-\left(Axe^{-x}+Bx^2+Cx+D\right)=2e^{-x}+3x^2$$

$$-2Ae^{-x}-Bx^2-Cx-D+2B=2e^{-x}+3x^2+0x+0$$

Equating coefficients, we obtain:

$$-2A=2\implies A=-1$$

$$-B=3\implies B=-3$$

$$C=0$$

$$-D+2B=0\implies D=-6$$

And so our particular solution is:

$$y_p(x)=-xe^{-x}-3x^2-6$$

And thus, the solution to the ODE is:

$$y(x)=y_h(x)+y_p(x)=c_1e^{x}+c_2e^{-x}-xe^{-x}-3x^2-6$$

 

[karush]

probably

however the mml just closed so I can't check.
but the rewrite you gave is probably the one I submitted that was correct
unfortunately I got the answer from W|A
but didn't know the steps

wow, thanks again I would of spent a lot of time trying to get it

 

[HallsofIvy]

ok going to see if i can do the steps
$\textrm{17.2.9 Solve the given equation by the method of undetermined coefficients.}\\$
\begin{align*}\displaystyle
y''-y&=2e^{-x}+3e^{2}
\end{align*}
$\textit{ textbook answer is:}\\$
\begin{align*}\displaystyle
y&=c_1e^{-x}+c_2 e^{-x}-xe^{-x}-3x^{2}-6\\
\end{align*}

That "text book answer" very clearly is NOT correct. For one thing, $$c_1e^{-x}+ c_2e^{-x}= (c_1+ c_2)e^{-x}= ce^{-x}$$. I presume you meant $$c_1e^x+ c_2e^{-x}$$. But even $$y= c_1e^x+ c_2e^{-x}- xe^{-x}- 3x^2- 6$$ does NOT satisfy the given equation. For that y, $$y'= c_1e^x- c_2e^{-x}- e^{-x}+ xe^{-x}- 6x$$ and $$y''= c_1e^x+ c_2e^{-x}+ 2e^{-x}- xe^{-x}- 6$$ so that $$y''- y= 2e^{-x}- xe^{-x}- 6$$ NOT $$2e^{-x}+ 3e^2$$.