What is the general solution of the differential equation? : $(x - 4) y^{4} d x - x^{3} (y^{2} - 3) d y = 0$ $(x - 4) y^4 dx - x^3 (y^2 - 3) dy =0$

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Answer 1 · 2017-07-09T22:08:00

$\frac{1}{y^{3}} - \frac{1}{y} = \frac{2}{x^{2}} - \frac{1}{x} + c$ $1/y^3 -1/y = 2/x^2 -1/x+ c$

We have the following Differential Equation in differential form

$(x - 4) y^{4} d x - x^{3} (y^{2} - 3) d y = 0$ $(x - 4) y^4 dx - x^3 (y^2 - 3) dy =0$

Which we can re-arrange as follows:

$\frac{y^{2} - 3}{y^{4}} \frac{d y}{d x} = \frac{x - 4}{x^{3}}$ $(y^2 - 3)/y^4 dy/dx = (x - 4)/x^3$

This is now a separable DIfferential Equation, and so "separating the variables" gives us:

$int \ (y^2 - 3)/y^4 \ dy = int \ (x - 4)/x^3 \ dx$

This is now a trivial integration problem, thus:

$int \ 1/y^2 - 3/y^4 \ dy = int \ 1/x^2 - 4/x^3 \ dx$

$:. -1/y + 1/y^3 = -1/x + 2/x^2 + c$

Hence the solution is:

$1/y^3 -1/y = 2/x^2 -1/x+ c$

What is the general solution of the differential equation? : (x - 4) y^4 dx - x^3 (y^2 - 3) dy =0 (x−4)y4dx−x3(y2−3)dy=0 (x - 4) y^4 dx - x^3 (y^2 - 3) dy =0