Solve separation of variables problem

Question

Originally I had $\frac{d^2y}{dt^2}=-A e^{y/B} (\frac{dy}{dt})^2$.

Using a given hint: $\frac{dx}{dy}=\frac{dx}{dt}\frac{dt}{dy}=\frac{d^2y}{dt^2}\frac{1}{x}$ and $x=\frac{dy}{dt}$

I got: $x\frac{dx}{dy}=-A e^{y/B} x^2$, separation of variables gave:

$\frac{1}{x} dx = -A e^{y/B} dy$

Where $A,B$ are constants. The initial condition is $x_0=x$. Also when solving, let $y=-\infty$ Now I'm a little stuck.

Answer 1

Hint: $$\frac{d(x^2)}{dy} = 2x\frac{dx}{dy}.$$

And in general form, if you have an equation $x' = f(x)g(t)$ with initial data $x(t_0) = x_0$, then the method of separation of variables writes

$$\int_{x_0}^{x(t)}\frac{du}{f(u)}=\int_{t_0}^{t}g(s)ds.$$ After that you study the existence of these integrals, find them, and try to solve the result with respect to $x(t)$.

If you have any further questions, ask in comments.

EDIT

Ok, let's write it down properly.

$$\int_{x_0}^{x(y)}\frac{du}{u}=-\int_{y_0}^{y}Ae^{s/B}ds,$$ $$\ln (x(y)/x_0)=-AB(e^{y/B}-e^{y_0/B}).$$ Therefore $$ x(y)=-x_0e^{AB(e^{y/B}-e^{y_0/B})}.$$

Suppose that $B>0$ and $y_0=-\infty$. Then we can say that $x(y) =-x_0e^{AB e^{y/B} } $.

Answer 2

I believe you are on the right track. Starting with $$ \frac{1}{x} dx = -A e^{y/B} dy $$ you have the equation separated. Now you can integrate each side to get $$ \ln(x)=-ABe^{y/B} + C, $$ where $C$ is an integration constant to be determined with the boundary condition, e.g. $y(x=0)=y_{0}$. With some algebra you can write the solution for $y$ explicitly as $$ y(x) = B \cdot \ln[\frac{\ln(x)+C}{-AB}].$$

Hope this helps some,

Paul Safier

Or, if you want the equation written explicitly in $x$ with the boundary condition of $x(y=0)=x_{0}$ you'll find that $C=\ln(x_{0})+AB$, and thus, $$ x(y)=\exp[C-ABe^{y/B}],$$ where $$C=\ln(x_{0})+AB.$$

Cheers,

Paul Safier