# How to solve $x'(t)=\frac{x+t}{2t-x}$?

I wish to solve $x'(t)=\frac{x+t}{2t-x}$ with the initial condition $x(1)=0$. I noted that $x'(t)=f(\frac{x}{t})$ where $f(y)=\frac{y+1}{2-y}$ so I denoted $y(t)=\frac{x(t)}{t}$ and got that $y'(t)=\frac{f(y)-y}{t}$ so I can write something like $\frac{dy}{dt}=\frac{f(y)-y}{t}$ so $\frac{dy}{f(y)-y}=\frac{dt}{t}$ .

Here I am a bit stuck, I know I should do something like take integrals on both sides, but I am having trouble with the initial condition. In this exercise I can leave integrals in the answer so I would like to know how to get the solution with the integral, I think this requires to calculate the boundaries of some integral (maybe of $\frac{f(y)-y}{t}$ ?)

How can I continue to get the solution with integral that satisfies the initial condition ? Help is appriciated!

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Your differential equation is homogenous of order one. You did it right finding $\frac{dy}{dt}=\frac{f(y)-y}{t}$. So you can integrate of both sides in last equation till your solution is achieved respect to $y(t)$ and $t$. Then you can substitute $x(t)=y(t)t$ in your solution and then apply the initial condition. I hope it help. :)