Integration Help

[Peregrine]

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.

[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

[HallsofIvy]

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}
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
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.

And trying to integrate again, Using u-substitution, [tex]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?
No, you cannot leave "y" in something you are integrating with respect to u.

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

[Peregrine]

Thanks for the help!