Question #b0376

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Answer 1 · 2018-01-20T03:38:01

$(xy)/(x^2-y^2)^2= c_1$

Let $y=vx$ , where v is a function of x, which means $dy/dx= v +x (dv)/dx$ . The given DE , thus transforms to $v+x(dv)/dx =(v^3 +3v)/(1+3v^2)$

$x(dv)/dx = (v^3+3v)/(1+3v^2) -v =(2v-2v^3)/(1+3v^2)$

$(1+3v^2)/(v(1-v^2))dv= 2(1/x) dx$ . Partial fractions on the left side would be

$(1/v +(4v)/(1-v^2))dv= 2 (1/x) dx$ . Now integrate on both sides to get

$ln v -2ln(1-v^2) = 2 lnx +C$ . On simplification it becomes,

$ln (v/(1-v^2)^2) = ln x^2 +C$
$v/(1-v^2)^2 = c_1x^2$

$(yx^3)/(x^2-y^2)^2= c_1x^2$

$(xy)/(x^2-y^2)^2 =c_1$