Solve #(y−4x−1)^2dx−dy=0#?

Question

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

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Answer 1 · 2018-01-18T21:31:20

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

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#