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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!

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    $\begingroup$ No, can't be good as the only $x^2$ term in the LHS is $x^2$, while it is $2x^2$ in the RHS. Easier to start with $y=x^2+Ax+B$. $\endgroup$ – Yves Daoust Sep 17 '14 at 13:33
  • $\begingroup$ The second derivative should be typed as y'', not as y". (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. $\endgroup$ – user147263 Sep 20 '14 at 6:23
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The equation is $$ (D^{2}-4D+2)y=2x^{2} $$ Apply the annihilator $D^{3}$ of $x^{2}$ to obtain $$ D^{3}(D^{2}-4D+2)y=0. $$ The general solution of this equation is $$ f=e^{2x}(Ae^{\sqrt{2}x}+Be^{-\sqrt{2}x})+F+Gx+Hx^{2}. $$ Plugging back into the original equation eliminates the exponential terms and gives $$ \begin{align} (D^{2}-4D+2)f & =(D^{2}-4D+2)(F+Gx+Hx^{2}) \\ & =(2H)-4(G+2Hx)+2(F+Gx+Hx^{2}) \\ & =(2H-4G+2F)+(-8H+2G)x+2Hx^{2}= x^{2}. \end{align} $$ Therefore $$ \begin{array}{l} H=1/2\\ -8H+2G=0 \implies G=4H=2 \\ 2H-4G+2F=0 \implies F=2G-H=4-1/2=7/2. \end{array} $$ The final solution is $$ f = e^{2x}(Ae^{\sqrt{2}x}+Be^{-\sqrt{2}x})+\frac{7}{2}+2x+\frac{1}{2}x^{2}. $$ $A$ and $B$ are arbitrary constants.

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Your steps look good. If you want to check the solution, plug the solution and its derivatives back into the original equation to see if it works.

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Your $y_c$ part is fine but there is an error in $y_p$ part. Coefficients should be $A=1$, $B=4$ and $C=7$.

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