$$x+ydydxy$$−$$xdydx=x2+2y2+y4x2$$ , the solution of the differential equation is kyx−$$1x2+y2=C$$, then $$k=$$

Given equation can be rewritten $$asxdx+ydyydx$$−$$xdy=x4+2x2y2+y4x2$$⇒$$xdx+ydyx2+y22=ydx$$−$$xdyy2$$×$$y2x2$$⇒∫$$dx2+y2x2+y22=2$$∫$$1(x/y)2$$⋅dxy⇒−$$1x2+y2=$$−$$2x/y+c$$⇒$$2yx$$−$$1x2+y2=C$$

Methods of solving first order first degree differential equations

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Question

$\frac{x + y \frac{d y}{d x}}{y - x \frac{d y}{d x}} = x^{2} + 2 y^{2} + \frac{y^{4}}{x^{2}}$ , $t h e s o l u t i o n o f t h e d i f f e r e n t i a l e q u a t i o n i s$ $\frac{k y}{x} - \frac{1}{x^{2} + y^{2}} = C, t h e n k =$

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Solution

Given equation can be rewritten as
$\begin{array}{l} \frac{x d x + y d y}{y d x - x d y} = \frac{x^{4} + 2 x^{2} y^{2} + y^{4}}{x^{2}} \\ \Rightarrow \frac{x d x + y d y}{{(x^{2} + y^{2})}^{2}} = \frac{y d x - x d y}{y^{2}} \times \frac{y^{2}}{x^{2}} \\ \Rightarrow \int \frac{d (x^{2} + y^{2})}{{(x^{2} + y^{2})}^{2}} = 2 \int \frac{1}{(x / y)^{2}} \cdot d (\frac{x}{y}) \\ \Rightarrow \frac{- 1}{x^{2} + y^{2}} = \frac{- 2}{x / y} + c \\ \Rightarrow \frac{2 y}{x} - \frac{1}{x^{2} + y^{2}} = C \end{array}$

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Methods of solving first order first degree differential equations

Remember concepts with our Masterclasses.

x+ydydxy−xdydx=x2+2y2+y4x2 , the solution of the differential equation is kyx−1x2+y2=C, then k=

Given equation can be rewritten asxdx+ydyydx−xdy=x4+2x2y2+y4x2⇒xdx+ydyx2+y22=ydx−xdyy2×y2x2⇒∫dx2+y2x2+y22=2∫1(x/y)2⋅dxy⇒−1x2+y2=−2x/y+c⇒2yx−1x2+y2=C

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$\frac{x + y \frac{d y}{d x}}{y - x \frac{d y}{d x}} = x^{2} + 2 y^{2} + \frac{y^{4}}{x^{2}}$ , $t h e s o l u t i o n o f t h e d i f f e r e n t i a l e q u a t i o n i s$ $\frac{k y}{x} - \frac{1}{x^{2} + y^{2}} = C, t h e n k =$