# Unable to find a suitable transformation

#### captainvhagar

##### New member

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

#### Subhotosh Khan

##### Super Moderator
Staff member
View 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
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.

#### HallsofIvy

##### Elite Member
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.