What is the general solution of the differential equation? : $x y \frac{d y}{d x} = x^{2} - y^{2}$ $xy dy/dx = x^2 - y^2$

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Answer 1 · 2017-09-22T12:57:29

$y = \pm \sqrt{\frac{x^{2}}{2} - \frac{A}{x^{2}}}$ $y = +-sqrt(x^2/2 - A/x^2)$

We have:

$x y \frac{d y}{d x} = x^{2} - y^{2}$ $xy dy/dx = x^2 - y^2$ ..... [A]

Note, that we cannot isolate terms in $x$ $x$ and $y$ $y$ alone so let us attempt some manipulation in an attempt to simplify the equation:

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

$:. dy/dx = (x^2)/(xy) - (y^2)/(xy)$

$:. dy/dx = x/y - y/x$ ..... [B]

So Let us try a substitution, Let:

$v = y/x => y=vx => dy/dx = v + x(dv)/dx$

And substituting into the above DE [B], to eliminate $y$ :

$v + x(dv)/dx = 1/v - v$
$:. x(dv)/dx = 1/v - 2v$

$:. x(dv)/dx = (1 - 2v^2)/v$

$:. v/(1-2v^2) (dv)/dx = 1/x$

Which has indeed helped as it has reduced the initial equation [A] to a separable equation, so we can now "seperate the variables" to get:

$int \ v/(1 - 2v^2) \ dv = int \ 1/x \ dx$
$:. -1/4 \ int \ (-4v)/(1 - 2v^2) \ dv = int \ 1/x \ dx$

And this is trivial to integrate to get:

$-1/4 ln | 1 - 2v^2 | = ln |x| + lnC$
$:. -1/4 ln | 1 - 2v^2 | = ln |Cx|$
$:. ln | 1 - 2v^2 | = (-4)ln |Cx|$

$:. ln | 1 - 2v^2 | = ln |1/(Cx)^4|$

And the due to the monotonicity of the logarithmic function we have:

$1 - 2v^2 = 1/(Cx)^4$

And restoring the substitution we get:

$1 - 2(y/x)^2 = 1/(Cx)^4$

Which is the General Solution of [A], and is we require an explicit solution; then:

$2(y/x)^2 = 1 - 1/(Cx)^4$
$:. y^2/x^2 = 1/2 - 1/2 1/(Cx)^4$

$:. y^2 = x^2{ 1/2 - 1/2 1/(Cx)^4 }$

$:. y^2 = x^2/2 - A/x^2$

Leading to the explicit solution:

$y = +-sqrt(x^2/2 - A/x^2)$

What is the general solution of the differential equation? : xy dy/dx = x^2 - y^2 xydydx=x2−y2 xy dy/dx = x^2 - y^2