# How to find verify solution to differential equation

1. Sep 3, 2012

### digipony

1. The problem statement, all variables and given/known data

See attached photo

2. Relevant equations

3. The attempt at a solution
I don't even know where to start on this problem. Usually, if I am given a possible solution to a DE, say y(t)=cos(t) as a solution to y''-y'+y=sin(t), I find whether it was a solution or not by manipulating it and then plugging it into the second equation that it may be a solution to and then seeing whether the resulting statement is true. For the ex. problem I give in this section, I would do this:
y'= -sint
y''= -cost
then sub them in: -cost +sint +cost = sint
and by simplifying get: sint = sint
so for this case, it y=cost is a solution.
But with the problem that is in the attached photo I have no idea what to do.

Does anyone have an guidance as to how I can solve the attached problem?
Thanks :)

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2. Sep 3, 2012

### SammyS

Staff Emeritus
Here's the "photo":

Well, find $\displaystyle \frac{\partial}{\partial x} f(x-ct)$ and then the partial with respect to x of that.

Do the same with partials with respect to t.

Then plug all that in.

3. Sep 3, 2012

### digipony

I don't have much experience with partial derivatives, am I on the right track here: the first partial derivative with respect to x, $$\displaystyle\frac{\partial}{\partial x} f(x-ct)$$ be: 1, therefore the second would be 0?

4. Sep 3, 2012

### digipony

Sorry for the formatting on that, I had not even heard of LateX before finding this site.
Oh never mind this, it's fixed.

Last edited: Sep 3, 2012
5. Sep 3, 2012

### SammyS

Staff Emeritus
No.

$\displaystyle \frac{\partial}{\partial x} f(x-ct) = (1)\cdot f'(x-ct)\,,$ because d/dx (x) = 1 .

Similarly $\displaystyle \frac{\partial}{\partial t} f(x-ct) = (-c)\cdot f'(x-ct)\,.$

6. Sep 3, 2012

### digipony

Oh, so that's what the reference to chain rule was about! Thank you!!