Solve (y−4x−1)^2dx−dy=0?

I need to find a general solution to this ODE: (y−4x−1)^2dx−dy=0

1 Answer
Jan 18, 2018

y = 4/(Ce^(4x)+1) +4x -1

Explanation:

Given: (y−4x−1)^2dx−dy=0

dy/dx = (y-4x-1)^2

Let u = y-4x, then (du)/dx = dy/dx -4

(du)/dx+4 = (u-1)^2

(du)/dx = u^2-2u-3

The equation is separable:

(du)/(u^2-2u -3) = dx

int(du)/(u^2-2u -3) = intdx

I used WolframAlpha for the integration of the left side:

ln(((3-u)/(u+1))^(1/4)) = x + C

(3-u)/(u+1) = Ce^(4x)

-1 + 4/(u+1) = Ce^(4x)

4/(u+1) = Ce^(4x)+1

u+1 = 4/(Ce^(4x)+1)

y = 4/(Ce^(4x)+1) +4x -1