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[SOLVED] 17.2.9 Solve the given equation by the method of undetermined coefficients

karush

Well-known member
Jan 31, 2012
2,928
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

Administrator
Staff member
Feb 24, 2012
13,775
Could if be that the given ODE is:

\(\displaystyle y''-y=2e^{-x}+3x^2\)

And the textbook answer is equivalent to:

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I will proceed as if what I posted is correct, and give my solution below:

We see that the characteristic roots are:

\(\displaystyle r=\pm1\)

And so the homogeneous solution is:

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

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

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

Differentiating, we find:

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

Substitution into the ODE gives us:

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

\(\displaystyle -2Ae^{-x}-Bx^2-Cx-D+2B=2e^{-x}+3x^2+0x+0\)

Equating coefficients, we obtain:

\(\displaystyle -2A=2\implies A=-1\)

\(\displaystyle -B=3\implies B=-3\)

\(\displaystyle C=0\)

\(\displaystyle -D+2B=0\implies D=-6\)

And so our particular solution is:

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

And thus, the solution to the ODE is:

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

karush

Well-known member
Jan 31, 2012
2,928
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:confused:

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

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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, [tex]c_1e^{-x}+ c_2e^{-x}= (c_1+ c_2)e^{-x}= ce^{-x}[/tex]. I presume you meant [tex]c_1e^x+ c_2e^{-x}[/tex]. But even [tex]y= c_1e^x+ c_2e^{-x}- xe^{-x}- 3x^2- 6[/tex] does NOT satisfy the given equation. For that y, [tex]y'= c_1e^x- c_2e^{-x}- e^{-x}+ xe^{-x}- 6x[/tex] and [tex]y''= c_1e^x+ c_2e^{-x}+ 2e^{-x}- xe^{-x}- 6[/tex] so that [tex]y''- y= 2e^{-x}- xe^{-x}- 6[/tex] NOT [tex]2e^{-x}+ 3e^2[/tex].