# solve by using variation of parameters

by SOS2012
Tags: complementary, differential, general solution, variation parameters
 P: 4 x²y"(x)-3xy'(x)+3y(x)=2(x^4)(e^x) =>y"(x)-(3/x)y'(x)+(3/x²)y(x)=2x²e^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!!!
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,898 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. $$\frac{dy}{dx}= \frac{dy}{dt}\frac{dt}{dx}= \frac{1}{x}\frac{dy}{dt}$$ and, differentiating again, $$\frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{dy}{dx}\right)$$$$= \frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dt}\right)= -\frac{1}{x^2}\frac{dy}{dt}+ \frac{1}{x^2}\frac{d^2y}{dt^2}$$
P: 4
 Quote by HallsofIvy 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. $$\frac{dy}{dx}= \frac{dy}{dt}\frac{dt}{dx}= \frac{1}{x}\frac{dy}{dt}$$ and, differentiating again, $$\frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{dy}{dx}\right)$$$$= \frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dt}\right)= -\frac{1}{x^2}\frac{dy}{dt}+ \frac{1}{x^2}\frac{d^2y}{dt^2}$$
thank you very much. i appreciate the help

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