Partial fractions (?) to solve first order DE

[huckleberry02]

hello world,

I've been doing some summertime training to brush up my math skills and have been struggling with this

[dy]/[/dt]=(4exp(-y)+const*exp(-2y))^1/2

In fact this is the simplified version of a Bernouilli equation. I know that it is separable, I'm just struggling with the integration with respect to "y".

If anybody could help that'd be wonderful. Thanks a lot, have a nice day!

~huckleberry

 

[HallsofIvy]

Your equation is
$$\frac{dy}{dx}= \left(e^{-y}+ Ce^{-2y}\right)^{1/2}$$
which can be written as
$$\frac{dy}{\left(e^{-y}+ Ce^{-2y}\right)^{1/2}}= dx$$

I would let $u= e^{-y}$ so that $du= -e^{-y}dy$ and $du/u= -dy$ In terms of u, the equation becomes
$$\frac{du}{u\left(u+ Cu^2\right)^{1/2}}= dt$$

 

[huckleberry02]

HallsOfIvy,

Thank you for you response.

I tried your method and I've run up against a similar problem as before, when i tried partial fractions. I can't seem to solve correctly for the coefficients.

A/u +B/(u+C*u^2)^(1/2)=1

A*(u+C*u^2)^(1/2) +Bu=1

choosing u=(-1/C) --> B=-C

however, I get stuck looking A, the only way to make the B drop is to set u=0, which is not coherent.

do you have a hint for me? thanks.

~huckleberry

 

[huckleberry02]

ps ( how do you enter the equations so nicely?)

 

[haruspex]

I don't believe you can use partial fractions when there are surds involved.
Try making a change of variable to get the surd in the form √(u²+B) (where B may be negative in this case), then look for a trig or hyperbolic trig substitution to make the surd collapse.
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[arildno]

At some point, after completing the square and having eliminated the surd through haruspex's advice, you probably will need to make another substitution of te form v=Tan(u/2), or v=Tanh(u/2).