# Integration Help

1. Dec 10, 2006

### Peregrine

Simple Diff Eq Help

I am trying to solve the following Diff Eq:

$$\frac{d^2x}{dy^2}+(\frac{y}{2}-\frac{1}{y})\frac{dx}{dy}=0$$

I tried to solve by setting $$\frac{dx}{dy}=z$$

so: $$\frac{dz}{dy}+(\frac{y}{2}-\frac{1}{y})z=0$$

I know the general solution to this is:

$$z=-e^{-\int{\frac{y}{2}-\frac{1}{y}dy}}\int{0}dy$$

This then yields:
$$z=-C_1e^{ln(y)-1/4y^2}=\frac{dx}{dy}$$

And trying to integrate again, Using u-substitution, $$u = ln(y) -1/4y^2$$
$$du=(\frac{1}{y} - \frac{y}{2}) dy$$
$$dy = \frac{2}{2-y^2} du$$

Now, can I leave that y tern in the u-substitution? Or did I make a mistake along the way?

Also, is there an easier way to solve this integral than the path I've taken? Thanks.

Last edited: Dec 10, 2006
2. Dec 10, 2006

### CPL.Luke

I'd try not thinking about what the general solution should be and think about seperating the equation and solving from there.

however for you final integral, you need to symplify z a bit.

the integrand should be

Cye^-1/4y^2

which is easily integrable using the substitution u=y^2

3. Dec 11, 2006

### HallsofIvy

Staff Emeritus
That's valid, since this is a linear equation but I think it would be simpler to treat it as a separable equation:
$$\frac{dz}{z}= \left(\frac{1}{y}- \frac{y}{2}\right)dy[/itex] so [tex]ln(z)= ln(y)- \frac{y^2}{4}+ c$$
or
$$z= \frac{dx}{dy}= \frac{Cy}{e^{\frac{y^2}{4}}}$$
That also separates:
$$dx= \frac{Cy}{e^{\frac{y^2}{4}}}dy$$
To integrate that, let u= y2/4.

No, you cannot leave "y" in something you are integrating with respect to u.

Last edited: Dec 13, 2006
4. Dec 12, 2006

### Peregrine

Thanks for the help!