(x2+2xy−4y2)dx−(x2−8xy−4y2)dy=0

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grbavit · Accepted Answer

Since this is a homogeneous equation we put
$y = v x \Rightarrow d y = v d x + x d v$
Hence subtitiuting above we have
$(x^{2} + 2 x y - 4 y^{2}) d x - (x^{2} - 8 x y - 4 y^{2}) d y = 0$
$\Rightarrow (x^{2} + 2 v x^{2} - 4 v^{2} x^{2}) d x - (x^{2} - 8 v x^{2} - 4 v^{2} x^{2}) (x d v + v d x) = 0$
$\Rightarrow (x^{2} + 2 v x^{2} - 4 v^{2} x^{2}) d x - (x^{3} - 8 v x^{3} - 4 v^{2} x^{3}) d v - (v x^{2} - 8 v^{2} x^{2} - 4 v^{3} x^{2}) d x = 0$
deviding both sides by $x^{2}$
$\Rightarrow (1 + 2 v - 4 v^{2}) d x - (x - 8 v x - 4 v^{2} x) d v - (v - 8 v^{2} - 4 v^{3}) d x = 0$
$\Rightarrow (1 + 2 v - 4 v^{2} - v + 8 v^{2} + 4 v^{3}) d x - x (1 - 8 v - 4 v^{2}) d v = 0$
$\Rightarrow (1 + v + 4 v^{2} + 4 v^{3}) d x = x (1 - 8 v - 4 v^{2}) d v$
$\Rightarrow \frac{d x}{x} = \frac{1 - 8 v - 4 v^{2}}{1 + v + 4 v^{2} + 4 v^{3}} d v$
$\Rightarrow \frac{d x}{x} = \frac{1 - 8 v - 4 v^{2}}{(1 + v) (1 + 4 v^{2})} d v$
$\Rightarrow \frac{d x}{x} = \frac{(4 u^{2} + 1) - 8 u (u + 1)}{(1 + v) (1 + 4 v^{2})}$
$\Rightarrow \frac{d x}{x} = [\frac{4 u^{2} + 1}{(1 + v) (1 + 4 v^{2})} - \frac{8 u (u + 1)}{(1 + v) (1 + 4 v^{2})}] d v$
$\Rightarrow \frac{d x}{x} = [\frac{1}{(1 + v)} - \frac{8 u}{(1 + 4 v^{2})}] d v$
$\Rightarrow \int \frac{d x}{x} = \int [\frac{1}{(1 + v)} - \frac{8 u}{(1 + 4 v^{2})}] d v + l n C$
$\Rightarrow \ln (x) = \ln (1 + v) - \ln (1 + 4 v^{2}) + \ln C$
$\Rightarrow x = C \frac{1 + v}{1 + 4 v^{2}}$
C is the integrating constant
putting back the value of y
$\Rightarrow x = C \frac{1 + \frac{y}{x}}{1 + 4 (\frac{y}{x})^{2}}$

(x^{2}+2xy-4y^{2})dx-(x^{2}-8xy-4y^{2})dy=0

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