# Homework Help: A diff. eq. or trigonometry related problem?

1. Mar 28, 2006

### parsifal

y' = 2y/x + x cos (y/x^2), with y=z*x^2
=> y' = 2zx + x cos z

and y=z*x^2 => y' = 2zx + x^2 * dz/dx

So that leaves x^2 * dz/dx = x cos z => dz/cos z = dx/x

I integrate both sides so that:

sec z + tan z = x + C

But I don't have a clue on how to get past that point. Should I start from the beginning with another technique to avoid getting sec z + tan z?

2. Mar 28, 2006

### HallsofIvy

What exactly is the problem? What are you trying to do?

3. Mar 29, 2006

### parsifal

I'm trying to solve y, so I have to solve z in terms of x first and then substitute it in the equation y=z*x^2.

4. Mar 29, 2006

### HallsofIvy

You skipped over ln parts and have a minor error: the right side should be, after taking exponentials of both sides, Cx, not x+ C.

Also, rather than using the "Table" formula
$$\int sec z dz= ln|sec z+ tan z|+ C[/itex] I did it as [tex]\int \frac{dz}{cos z}= \int \frac{cos z dz}{cos^2 z}= \int \frac{cos z dz}{1- sin^2 z}$$
Now the substitution u= sin z gives a rational integral that can be integrated by partial fractions. The result is, of course, similar to the above but in terms of sin z only and so easier to solve for z.

5. Mar 29, 2006

### parsifal

Ok, I think I can carry on from that.

Thank you!

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