differential equation using method of undetermined coefficients

**twilightstr** · May 4th 2009, 06:45 PM

y'' + y = e^x + x^3, y(0) = 2, y'(0)=0

**TheEmptySet** · May 4th 2009, 07:21 PM

Originally Posted by twilightstr

y'' + y = e^x + x^3, y(0) = 2, y'(0)=0

There are two parts to this problem.

First we need to find the complimentery solution(the solution to the homogenious solution)

$y''+y=0$ has the auxillary equation

$m^2+1=0 \iff m=\pm i$ so

$y_c=c_1\cos(t)+c_2\sin(t)$

Now we need to find the particular solution

$y_p=Ae^{x}+B+Cx+Dx^2+Ex^3$

$y_p'=Ae^{x}+C+2Dx+3Ex^2$

$y_p''=Ae^{x}+2D+6Ex$

Plugging these in we get

$Ae^{x}+B+Cx+Dx^2+Ex^3+Ae^{x}+2D+6Ex=e^{x}+x^3$

$2Ae^{x} +Ex^3+Dx^2 +(6E+C)x+(2D+B)=e^{x}+x^3$

From this we see that both $D=0,B=0$

$2A=1 \iff A=\frac{1}{2}$

$E=1$ and

$6E+C=0 \iff 6+C=0 \iff C=-6$

$y_p=\frac{1}{2}e^{x}+x^3-6x$

So the solution by superposition

$y=y_c+y_p=c_1\cos(t)+c_2\sin(t)+\frac{1}{2}e^{x}+x ^3-6x$

From here just plug in you I.C and you are done

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