Question

What is a solution to the differential equation `y'=3^-y(2x-4)` and y(5)=0?

Answer 1

`y =( ln (ln 3 (x^2-4x - 5) + 1) )/ ln 3`

Explanation:

`y'=3^-y(2x-4)`

this is separable

`3^y \ y'=2x-4`

`int \ 3^y \ y' \ dx=int \ 2x-4 \ dx`

we can explore the integral on the LHS a little

consider `u = C^v`
`ln u = v ln C`
`1/u (du)/(dv) = ln C`
`implies d/(dv)(C^v) = ln C \ C^v`
so `d/dx (3^y) = ln 3 \ 3^y \ y'`

and so the integral is
`int \d/dx( 1/(ln 3) 3^y ) \ dx=int \ 2x-4 \ dx`

`1/(ln 3) 3^y =x^2-4x + C`

`3^y =ln 3 (x^2-4x) + C`

`y ln 3 = ln (ln 3 (x^2-4x) + C )`

`y =( ln (ln 3 (x^2-4x) + C ))/ ln 3`

applying the IV: `y(5) = 0`

`0 =( ln (ln 3 (25-20) + C ))/ ln 3`

`implies ln 3 (5) + C = 1`

`y =( ln (ln 3 (x^2-4x - 5) + 1) )/ ln 3`