# Initial value problem with two equations

I'm given the IVP below:

$$\frac{dx}{dt}=\frac{y}{y+x}+\ln(x+y)$$

$$\frac{dy}{dt}=-\frac{y}{y+x},$$

$$y(0)=e, x(0)=0$$

I started by dividing the two equation to get rid of the $$dt$$. I got:

$$\frac{dx}{dy}=-1+\frac{\ln(x+y)(x+y)}{-y}$$

I tried to substitute with $$u=x+y$$ but it din't help

Any thoughts?

• Adding the 2 equations and substituting $u=x+y$ gives $\frac{du}{dt} = \ln u$ which is separable, but the integral isn't elementary – Dylan Jan 27 at 17:28

Sum the equations to get $$\frac{d}{dt}(x+y)=\ln(x+y),\quad x+y\bigr|_{t=0}=e.$$ Solve it to get $$x+y$$, and then use the second equation to get $$y$$. The integrals coming up cannot be expressed in terms of elementary functions.