Solve Diff Eq: \frac{d^2x}{dy^2}+(\frac{y}{2}-\frac{1}{y})\frac{dx}{dy}=0

[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.

 

[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]<br /> so<br /> ln(z)= ln(y)- \frac{y^2}{4}+ c<br /> or<br /> z= \frac{dx}{dy}= \frac{Cy}{e^{\frac{y^2}{4}}}<br /> That also separates:<br /> dx= \frac{Cy}{e^{\frac{y^2}{4}}}dy<br /> To integrate that, let u= y²/4.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> And trying to integrate again, Using u-substitution, u = ln(y) -1/4y^2<br /> du=(\frac{1}{y} - \frac{y}{2}) dy<br /> dy = \frac{2}{2-y^2} du<br /> <br /> Now, can I leave that y tern in the u-substitution? Or did I make a mistake along the way? </div> </div> </blockquote> No, you cannot leave &quot;y&quot; in something you are integrating with respect to u.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Also, is there an easier way to solve this integral than the path I&#039;ve taken? Thanks. </div> </div> </blockquote>

 

[Peregrine]

Thanks for the help!