Find a particular solution of the differential equation `(x+1)(dy)/(dx)=2e^(-y)-1`given that `y = 0`when`x = 0`.

Explanation

Given,

`=>((x+1)dy​)/(dx)=2e^(−y)−1`

`=>(dy)/(2e^(−y)−1)=(dx)/(x+1)`

`=>(dy)/(2e^(−y)−1)=ln(x+1)+c`

Let `t=2e^(−y)−1`

So, `dt=−2e^(−y)dy`

So,from equation,

`=>(-dt)/(t(t+1))=ln(x+1)+c`

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