Solve ${(y - 4 x - 1)}^{2} d x - d y = 0$ $(y−4x−1)^2dx−dy=0$ ?

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

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Answer 1 · 2018-01-18T21:31:20

$y = \frac{4}{C e^{4 x} + 1} + 4 x - 1$ $y = 4/(Ce^(4x)+1) +4x -1$

Given: ${(y - 4 x - 1)}^{2} d x - d y = 0$ $(y−4x−1)^2dx−dy=0$

$\frac{d y}{d x} = {(y - 4 x - 1)}^{2}$ $dy/dx = (y-4x-1)^2$

Let $u = y - 4 x$ $u = y-4x$ , then $\frac{d u}{d x} = \frac{d y}{d x} - 4$ $(du)/dx = dy/dx -4$

$\frac{d u}{d x} + 4 = {(u - 1)}^{2}$ $(du)/dx+4 = (u-1)^2$

$\frac{d u}{d x} = u^{2} - 2 u - 3$ $(du)/dx = u^2-2u-3$

The equation is separable:

$\frac{d u}{u^{2} - 2 u - 3} = d x$ $(du)/(u^2-2u -3) = dx$

$\int \frac{d u}{u^{2} - 2 u - 3} = \int d x$ $int(du)/(u^2-2u -3) = intdx$

I used WolframAlpha for the integration of the left side:

$ln ({(\frac{3 - u}{u + 1})}^{\frac{1}{4}}) = x + C$ $ln(((3-u)/(u+1))^(1/4)) = x + C$

$\frac{3 - u}{u + 1} = C e^{4 x}$ $(3-u)/(u+1) = Ce^(4x)$

$- 1 + \frac{4}{u + 1} = C e^{4 x}$ $-1 + 4/(u+1) = Ce^(4x)$

$\frac{4}{u + 1} = C e^{4 x} + 1$ $4/(u+1) = Ce^(4x)+1$

$u + 1 = \frac{4}{C e^{4 x} + 1}$ $u+1 = 4/(Ce^(4x)+1)$

$y = \frac{4}{C e^{4 x} + 1} + 4 x - 1$ $y = 4/(Ce^(4x)+1) +4x -1$