Question

How do you solve the differential equation `y'=e^(-y)(2x-4)`, where `y5)=0` ?

Answer 1

This is an example of a separable equation with an initial value.
`{dy}/{dx}=e^{-y}(2x-4)`
`y(5)=0`

by multiplying by `e^y` and by `dx`,
`Rightarrow e^ydy=(2x-4)dx`
by integrating,
`Rightarrow inte^ydy=int(2x-4)dx`
`Rightarrow e^y=x^2-4x+C`
by taking the natural log,
`Rightarrow y=ln(x^2-4x+C)`

Now, we need to find `C` using `y(5)=0`.
`y(5)=ln((5)^2-4(5)+C)=ln(5+C)=0`
`Rightarrow 5+C=1 Rightarrow C=-4`

Hence, the solution is `y=ln(x^2-4x-4)`.