Given, xydydx=x^2+2y^2 dydx=x^2+2y^2xy substitute y=vx→dydx=xdvdx+v xdvdx+v=x^2+2(vx)^2x(vx) xdvdx+v=1+2v^2v xdvdx=1+2v^2v v x ...

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Solving Homogeneous Differential Equations

solve xydyd...

Question

solve
$x y \frac{d y}{d x} = x^{2} + 2 y^{2}$

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Solution

Given,

$x y \frac{d y}{d x} = x^{2} + 2 y^{2}$

$\frac{d y}{d x} = \frac{x^{2} + 2 y^{2}}{x y}$

substitute $y = v x \to \frac{d y}{d x} = x \frac{d v}{d x} + v$

$x \frac{d v}{d x} + v = \frac{x^{2} + 2 (v x)^{2}}{x (v x)}$

$x \frac{d v}{d x} + v = \frac{1 + 2 v^{2}}{v}$

$x \frac{d v}{d x} = \frac{1 + 2 v^{2}}{v} - v$

$x \frac{d v}{d x} = \frac{1 + v^{2}}{v}$

$\frac{v}{1 + v^{2}} d v = \frac{1}{x} d x$

$\frac{1}{2} log | 1 + v^{2} | = log | x | + c$
$\frac{1}{2} log | 1 + (y / x)^{2} | = log | x | + c$

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Similar questions

x y \frac{d y}{d x} = x^{2} + 2 y^{2}; y (1) = 0

\sqrt{1 + x^{2} + y^{2} + x^{2} y^{2}} + x y \frac{d y}{d x} = 0

x y \frac{d y}{d x} = \frac{1 + y^{2}}{1 + x^{2}} (1 + x + x^{2})

\sqrt{1 + x^{2} + y^{2} + x^{2} y^{2}} + x y \frac{d y}{d x} = 0

x y \frac{d y}{d x} = 2 y^{2} - x^{2}

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