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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?

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

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