Unable to find a suitable transformation

captainvhagar

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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
 
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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
You are claiming that the particular solution (yp) of this ODE is:

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

Did you check your "guess" by applying this solution to the ODE? I do not get the same yp.
 
Are you sure it is the dependent variable that is to be missing? Frankly, that doesn't make sense. This is an "Euler type" or "equipotent" (the exponent of x in each coefficient is equal to the order of the derivative) and there is a fairly standard method of converting that to a "constant coefficients" equation so there is no independent variable, x, does not appear in any coefficient.

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.
 
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