Solve the differential equation $x^2y_2=2y$

Question

Solve the differential equation $$x^2y_2=2y$$

My try:

Let $$z=x^2y_1$$

Differentiating with respect to $x$ we get

$$z_1=x^2y_2+2xy_1=x^2y_2+\frac{2z}{x}$$

$\implies$

$$z_1-\frac{2z}{x}=2y$$

So again differentiating with respect to $x$ we get

$$z_2-\frac{2z_1}{x}+\frac{2z}{x^2}=2y_1=\frac{2z}{x^2}$$

Hence

$$z_2=\frac{2}{x}z_1$$

Letting $z_1=p$

we have

$$\frac{dp}{dx}=\frac{2p}{x}$$

So we get

$$p=Cx^2$$

$$\frac{dz}{dx}=Cx^2$$

$$z=\frac{Cx^3}{3}+D$$

$$x^2y_1=\frac{Cx^3}{3}+D$$

$$\frac{dy}{dx}=\frac{Cx}{3}+\frac{D}{x^2}$$

Hence

$$y=\frac{Cx^2}{6}-\frac{D}{x}+E$$

Where $C,D,E$ are arbitrary constants

But as the Order of D.E is $2$ why am i getting three constants?

Answer 1

HINT

If you plug your solution in the original DE, you will see it is not true for all values of $E$.

Answer 2

Another Hint

A differential equation is any equation containing one or more derivatives.

$$x^2y_2=2y$$

isn't a differential equation as it doesn't contain any derivatives.

The only thing that you could solve for $x^2y_2=2y$ would be finding $x,y,y_2$ in terms of the other two variables.

Answer 3

Assuming you mean in Newton's notation $$ x^2y''=2y, $$ this is a Cauchy-Euler equation where you can try to find basis solutions in the form $y=x^m$, which gives a characteristic equation in $m$, $$ 0=m^2-m-2=(m-2)(m+1) $$ so that the solution basis is indeed $x^{-1}, x^2$.

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You get an additional constant because you construct $z$ using the first derivative of $y$, so that the second order ODE you get for $z$ is actually a third order ODE for $y$. So it is natural that starting from this equation for $z$ and integrating up to $y$ you get 3 integration constants. You now have to insert this solution into the original ODE to get a condition on the three constants that leaves only 2 of them free.

$$ x^2(Cx^2+D/x+E)''=2Cx^2+2D/x\implies E=0. $$

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You could avoid that by using $u=xy$ as intermediate variable, then $$ xu''=x^2y''+2xy'=2(y+xy')=2(xy)'=2u' \\\implies u'=3Cx^2,~u=Cx^3+D \\\implies y=Cx^2+D/x $$