# Help Solving the following Differential equation

Jose Z

## Homework Statement

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

## 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 y1 is a solution, then another solution has the form Y(x)=v(x)y1
Then: Y'(x)=v'(x)y1+v(x)y1'
Y''(X)=v''(x)y1+v'(x)y1+v'(x)y1+v(x)y1'
Now, after replacing the y factors in (1) with Y and its derivatives, I get:
v''(x)y1+(2y1+p(x)y1=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+c1)/μ(x)=c1/(e-x/(x-1))=c1(ex(x-1))

Now, because z(x)=v'(x)→∫z(x)dx=v(x)
Then I have v(x)=c1∫(ex(x-1)dx=c1((x-2)ex+c2)
But, whenever I do Y(x)=v(x)y1, and replace it in (1), it doesn't satisfy the equation-

Can somebody help?

ingrammatical
In the first line I believe that you should have y''-(x/x-1)y'+y(1/(x-1))=0

Homework Helper
Welcome to PF;
Is y1 a solution?

When you are given a solution, always try reduction of order.
Try y(x)=v(x)e^x and solve for v.

Homework Helper
Dearly Missed

## Homework Statement

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

## 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 y1 is a solution, then another solution has the form Y(x)=v(x)y1
Then: Y'(x)=v'(x)y1+v(x)y1'
Y''(X)=v''(x)y1+v'(x)y1+v'(x)y1+v(x)y1'
Now, after replacing the y factors in (1) with Y and its derivatives, I get:
v''(x)y1+(2y1+p(x)y1=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+c1)/μ(x)=c1/(e-x/(x-1))=c1(ex(x-1))

Now, because z(x)=v'(x)→∫z(x)dx=v(x)
Then I have v(x)=c1∫(ex(x-1)dx=c1((x-2)ex+c2)
But, whenever I do Y(x)=v(x)y1, and replace it in (1), it doesn't satisfy the equation-

Can somebody help?

Please learn to use parentheses properly, so that you write ##\frac{1}{x-1}## instead of ##\frac{1}{x}-1##, as you have done throughout (thus making everything you write incorrect). It's easy: just write 1/(x-1) instead of 1/x-1.

Mentor
Please learn to use parentheses properly, so that you write ##\frac{1}{x-1}## instead of ##\frac{1}{x}-1##, as you have done throughout (thus making everything you write incorrect). It's easy: just write 1/(x-1) instead of 1/x-1.

I've tried solving it by multiplying everythin for (1/x-1), so my equation looks like:
@Jose Z, you have parentheses in what you wrote -- (1/x-1) -- the problem is that they are in the wrong place. They should be around the terms in the denominator, the way that Ray wrote the fraction, not around the entire fraction.