- #1

Jose Z

- 1

- 0

## Homework Statement

Solve (x-1)y''-xy'+y=0 , given x>1 and y1=e

^{x}

## Homework Equations

## The Attempt at a Solution

I've tried solving it by multiplying everythin for (1/x-1), so my equation looks like:

y''-(x/x-1)y'+y=0 (because x-1 is unequal to 0) (1)

so now the equation has the form of:

y''+p(x)y'+y=0, with p(x)=(-x/x-1).

Then, given the fact that y

_{1}is a solution, then another solution has the form Y(x)=v(x)y

_{1}

Then: Y'(x)=v'(x)y

_{1}+v(x)y

_{1}'

Y''(X)=v''(x)y

_{1}+v'(x)y

_{1}+v'(x)y

_{1}+v(x)y

_{1}'

Now, after replacing the y factors in (1) with Y and its derivatives, I get:

v''(x)y

_{1}+(2y

_{1}+p(x)y

_{1}=0,

Which can me solved as a first order differential equation by doing v''(x)=z'(x) and v'(x)=z(x).

Then, let μ(x)=e

^{∫p(x)dx}=e

^{∫(-x/x-1)dx}=e

^{-x-ln(x-1)}=e

^{-x}(1/x-1)

Then, z(x)=(∫μ(x)g(x)dx+c

_{1})/μ(x)=c

_{1}/(e

^{-x}/(x-1))=c

_{1}(e

^{x}(x-1))

Now, because z(x)=v'(x)→∫z(x)dx=v(x)

Then I have v(x)=c

_{1}∫(e

^{x}(x-1)dx=c

_{1}((x-2)e

^{x}+c

_{2})

But, whenever I do Y(x)=v(x)y

_{1}, and replace it in (1), it doesn't satisfy the equation-

Can somebody help?