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I have this differential equation : 4x^{2}y''+8x^{3}y'+(4x^{2}-3)y = 0 and it is given that the equation has a solution $y_{1}(x)=x^{r}.$

I set $y_2$ as $ y_{2}=x^{r}v(x).$

When I'm trying to find the first and second derivative I start to fail.

I get y'_2=x^{r-1}(xv'+rv) but the second is always long and makes no sense when I keep on with the problem. After I have put derivatives for the $y$, y' and y'' in the original equation I get complex equations where i see no way to use reduction of order.

Can someone please help me and tell me what I'm doing wrong.

2 Answers 2

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Let's assume that we have the following equation: y''+a(x)y'+b(x)y=0.

If we know that $y_1(x)$ is a solution and we search our general solution in the form of $y_2(x)=y_1(x)v(x)$, then we get the following:

y'_2=y'_1v+y_1v' and y''_2=y''_1v+y'_1v'+y'_1v'+y_1v''=y_1''v+2y'_1v'+y_1v''

Then by putting those values in the equation, we get

y''_1v+2y'_1v'+y_1v''+a(x)(y'_1v+y_1v')+b(x)y_1v=0

or

(y''_1+a(x)y'_1+b(x))v+(2y'_1+a(x)y_1)v'+y_1v''=0

As we know that $y_1$ is a solution therefore we have that

y''_1+a(x)y'_1+b(x)=0

So from our equation we get

(2y'_1+a(x)y_1)v'+y_1v''=0

v''+(2y'_1/y_1+a(x))v'=0

which is already order reducible.

Now, in our example by putting $y=x^r$ in the equation we get

$4r(r-1)x^r+8rx^{r+2}+4x^{r+2}-3x^r=0$ from where we get $r=-1/2$.

From here it follows that solving our equation is the same as solving the following equation:

y''+2xy'+y-3y/{4x^2}=0 => a(x)=2x. Thus by taking $y=x^rv$ we get

v''+(2(x^{-1/2})'/(x^{-1/2})+2x)v'=0 =>

v''+(x^{-1}+2x)v'=0=>

Take w=v' and get w'+(x^{-1}+2x)w=0 we are done.

I think the continuation is simple.

Sincerely,

Tigran

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It's probably easier if you first determine the (unique) value of $r$ which makes $y_1$ a solution.