#### captainvhagar

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- Thread starter captainvhagar
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You are claiming that the particular solution (yView attachment 13517

Just by inspection I can seem to see that 2/9*sqrt(x) is a solution. I can't seem to figure out what transformation would work to make the dependent variable disappear, though.

Thanks

\(\displaystyle y_p \ \ = \ \dfrac{2}{9} \sqrt{x} \)

Did you check your "guess" by applying this solution to the ODE? I do not get the same y

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Let t= ln(x). Then \(\displaystyle \frac{dy}{dx}= \frac{dy}{dt}\frac{dt}{dx}=\)\(\displaystyle \frac{1}{x}\frac{dy}{du}\) and \(\displaystyle \frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dt}\right)=\)\(\displaystyle -\frac{1}{x^2}\frac{dy}{dt}+ \frac{1}{x^2}\frac{d^2y}{dt^2}\). So \(\displaystyle x^2y''- 3xy'+ 4y=\)\(\displaystyle x^2\left(\frac{1}{x^2}\frac{d^2y}{dt^2}- \frac{1}{x}\frac{dy}{dt}\right)-\)\(\displaystyle 3x\left(\frac{1}{x}\frac{dy}{dt}\right)+ 4y\)\(\displaystyle = \frac{d^2y}{dt^2}- 4\frac{dy}{dt}+ 4y=\)\(\displaystyle x^{1/2}= e^{t/2}\).

The equation \(\displaystyle y''- 4y'+ 4y= e^{t/2}\) is easy to solve.