What is the general solution of the differential equation $2(y-4x^2)dx+xdy = 0$ ?

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Answer 1 · 2017-07-09T10:53:26

$y = 2x^2+c/x^2$

$2(y-4x^2)dx+xdy = 0$

Which we can re-arrange as follows:

$dy/dx = (-2 (y-4x^2))/x$
$" " = 8x-(2y)/x$

$:. dy/dx + (2y)/x = 8x$ ..... [A]

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

$dy/dx + P(x)y=Q(x)$

Then the integrating factor is given by;

$I = e^(int P(x) dx)$
$\ \ = exp(int \ 2/x \ dx)$
$\ \ = exp( 2lnx )$
$\ \ = exp( lnx^2 )$
$\ \ = x^2$

And if we multiply the DE [A] by this Integrating Factor, $I$ , we will have a perfect product differential;

$x^2 dy/dx + x^2xy = 8x^3$

$:. d/dx (x^2y) = 8x^3$

Which we can now directly integrate to get:

$x^2y = int \ 8x^3 \ dx$

$:. x^2y = 2x^4 + c$

$:. y = 2x^2+c/x^2$

What is the general solution of the differential equation 2(y-4x^2)dx+xdy = 0 2(y-4x^2)dx+xdy = 0 ?