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Solve by using variation of parameters

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SOS2012
#1
Nov19-12, 03:21 PM
P: 4
xy"(x)-3xy'(x)+3y(x)=2(x^4)(e^x)

=>y"(x)-(3/x)y'(x)+(3/x)y(x)=2xe^x

i dont know how to approach this problem because the coefficients are not constant and i am used to being given y1 and y2

HELP!!!
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HallsofIvy
#2
Nov19-12, 04:33 PM
Math
Emeritus
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Thanks
PF Gold
P: 39,497
This is an "Euler type" or "equi-potential" equation. The substitution t= ln(x) will change it to a "constant coefficients" problem in the variable t.

[tex]\frac{dy}{dx}= \frac{dy}{dt}\frac{dt}{dx}= \frac{1}{x}\frac{dy}{dt}[/tex]
and, differentiating again,
[tex]\frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{dy}{dx}\right)[/tex][tex]= \frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dt}\right)= -\frac{1}{x^2}\frac{dy}{dt}+ \frac{1}{x^2}\frac{d^2y}{dt^2}[/tex]
SOS2012
#3
Nov19-12, 04:41 PM
P: 4
Quote Quote by HallsofIvy View Post
This is an "Euler type" or "equi-potential" equation. The substitution t= ln(x) will change it to a "constant coefficients" problem in the variable t.

[tex]\frac{dy}{dx}= \frac{dy}{dt}\frac{dt}{dx}= \frac{1}{x}\frac{dy}{dt}[/tex]
and, differentiating again,
[tex]\frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{dy}{dx}\right)[/tex][tex]= \frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dt}\right)= -\frac{1}{x^2}\frac{dy}{dt}+ \frac{1}{x^2}\frac{d^2y}{dt^2}[/tex]
thank you very much. i appreciate the help


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