I'm trying to solving a 2nd-order inhomogeneous differential equation, but I'm not sure with my answer since I'm only learned it by myself & it has nothing to do with school or homework, so I have no one to ask or correct my answer. Please if you want to help me, i would be really happy & appreciate your help & advice.
So here's my problem
$y''\:-\:4y'\:+\:2y\:=\:2x^2$
and here's my step to answer
I know that to solve this 2nd-order inhomogeneous differential equation the general solution is
$y\:=\:y_c\:+\:y_p$
where $y_c$ is the general solution of the homogeneous equation. And $y_p$ is a particular solution
First, I find for $y_c$ through
$y''\:-\:4y'\:+\:2y\:=\:0$
since it's in the form of different roots of the auxiliary equation, i found that $m_1\:=\:2+\sqrt{2}$ and $m_2\:=\:2-\sqrt{2}$ , so the general solution of the homogeneous equation is
$y_c=c_1e^{\left(2+\sqrt{2}\right)x}+c_2e^{\left(2-\sqrt{2}\right)x}$
Second, since it's a polynomial problem so i find for $y_p$ through
$y_p=Ax^2+Bx+C$
first and second derivative of $y_p$ are
$y'_p=2Ax\:+\:B\:\:\:\:\:\:\:and\:\:\:\:\:\:\:\:y"_p=2A$
so now i have to find the undetermined coefficients $A$, $B$, and $C$. I find these coefficients through substitute $y_p$ into $y$, $y'_p$ into $y'$, and $y''_p$ into $y"$, thus
$y''\:-\:4y'\:+\:2y\:=\:\left(2A\right)x^2+\left(2B-8A\right)x\:+\:\left(2A-4B+2C\right)=2x^2$
So, $A\:=\:\frac{1}{2}\:,\:B\:=\:0,\:C\:=\:\frac{1}{2}$ and then substitute this into $y_p$, I find the particular solution which is
$y_p=\frac{1}{2}\left(x^2+1\right)$
and finally I find the general solution in the form of
$y\:=\:c_1e^{\left(2+\sqrt{2}\right)x}+c_2e^{\left(2-\sqrt{2}\right)x}+\frac{1}{2}\left(x^2+1\right)$
This is my answer, am I correct or not? Thank you for your time!
y''
, not asy"
. (Compare the outputs: $y''$ and $y"$.) I also suggest cutting down the spending on emoticons; the freed up funds can be used to buy some uppercase letters. $\endgroup$