# Help Solving the following Differential equation

• Jose Z
In summary, Jose Z's attempt at solving the homework equation resulted in multiplying every term for (1/x-1), but was unsuccessful because y1 was a solution. He then tried solving it by replacing the y terms with Y and its derivatives, but it did not satisfy the equation. He then found v(x) by solving c1(ex(x-1)) for x, which resulted in v(x)=c1(ex(x-1)).

## 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?

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

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.

Jose Z said:

## 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.

Ray Vickson said:
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.

Jose Z said:
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.

@Jose Z : How did you get on?

I feel we have been dumping on you somewhat ... I thought the use of parentheses in post #1 was clear from context and the proper use is only a recommended tweak for future consideration (It won't always be clear). As first posts go, yours was high quality: you have sincerely attempted the problem, as well as shown how you got stuck, so it is easy to decide what we need to tell you so you get the most out of these forums.

Well done and keep it up. :)