Question

x ( x - y ) dy = y ( x + y ) dx​

Answer 1

Answer:

$\sf -\dfrac{x}{2y} -\dfrac{1}{2}\: log|\dfrac{y}{x} |=log|x|+C$

Step-by-step explanation:

Given:

• The differential equation x (x - y) dy = y (x + y) dx

To Find:

• The solution for the given differential equation

Solution:

By given,

$\sf x\:(x-y\:)dy=y\:(x+y)\:dx$

$\sf \dfrac{dy}{dx} =\dfrac{y\:(x+y)}{x\:(x-y)}$

The given differential equation is homogeneous.

Hence,

Put y = vx

Differentiating on both sides with respect to x,

$\sf \dfrac{dy}{dx} =v+x\:\dfrac{dv}{dx}$

From the above equation,

$\sf v+x\:\dfrac{dv}{dx} =\dfrac{vx\:(x+vx)}{x\:(x-vx)}$

$\sf v+x\:\dfrac{dv}{dx} =\dfrac{v(x+vx)}{(x-vx)}$

Taking x common from both numerator and denominator,

$\sf v+x\:\dfrac{dv}{dx} =\dfrac{vx\:(1+v)}{x\:(1-v)}$

$\sf v+x\:\dfrac{dv}{dx} =\dfrac{v\:(1+v)}{(1-v)}$

$\sf v+x\:\dfrac{dv}{dx} =\dfrac{v+v^2}{(1-v)}$

$\sf x\:\dfrac{dv}{dx} =\dfrac{v+v^2}{(1-v)}-v$

$\sf x\:\dfrac{dv}{dx} =\dfrac{v+v^2-v+v^2}{(1-v)}$

$\sf x\:\dfrac{dv}{dx} =\dfrac{2v^2}{(1-v)}$

Now making it variable separable,

$\sf \dfrac{1-v}{2v^2} \: dv=\dfrac{1}{x}\: dx$

Integrate on both sides,

$\displaystyle \sf \int\limits {\dfrac{1-v}{2v^2} } \, dv =\int\limits {\dfrac{1}{x} } \, dx$

$\displaystyle \sf \int\limits {\dfrac{1}{2v^2} } \, dv -\int\limits {\dfrac{v}{2v^2} } \, dv =\int\limits {\dfrac{1}{x} } \, dx$

$\displaystyle \sf \dfrac{1}{2} \int\limits {\dfrac{1}{v^2} } \, dv -\dfrac{1}{2} \int\limits {\dfrac{1}{v} } \, dv =\int\limits {\dfrac{1}{x} } \, dx$

$\sf -\dfrac{1}{2v} -\dfrac{1}{2}\: log|v|=log|x|+C$

Now we know that y = vx

Hence v = y/x

Substitute the value of v,

$\sf -\dfrac{x}{2y} -\dfrac{1}{2}\: log|\dfrac{y}{x} |=log|x|+C$

This is the solution of the given differential equation.

Answer 2

Step-by-step explanation:

$Answer:</p><p></p><p>\sf -\dfrac{x}{2y} -\dfrac{1}{2}\: log|\dfrac{y}{x} |=log|x|+C−2yx−21log∣xy∣=log∣x∣+C</p><p></p><p>Step-by-step explanation:</p><p></p><p>Given:</p><p></p><p>The differential equation x (x - y) dy = y (x + y) dx</p><p></p><p>To Find:</p><p></p><p>The solution for the given differential equation</p><p></p><p>Solution:</p><p></p><p>By given,</p><p></p><p>\sf x\:(x-y\:)dy=y\:(x+y)\:dxx(x−y)dy=y(x+y)dx</p><p></p><p>\sf \dfrac{dy}{dx} =\dfrac{y\:(x+y)}{x\:(x-y)}dxdy=x(x−y)y(x+y)</p><p></p><p>The given differential equation is homogeneous.</p><p></p><p>Hence,</p><p></p><p>Put y = vx</p><p></p><p>Differentiating on both sides with respect to x,</p><p></p><p>\sf \dfrac{dy}{dx} =v+x\:\dfrac{dv}{dx}dxdy=v+xdxdv</p><p></p><p>From the above equation,</p><p></p><p>\sf v+x\:\dfrac{dv}{dx} =\dfrac{vx\:(x+vx)}{x\:(x-vx)}v+xdxdv=x(x−vx)vx(x+vx)</p><p></p><p>\sf v+x\:\dfrac{dv}{dx} =\dfrac{v(x+vx)}{(x-vx)}v+xdxdv=(x−vx)v(x+vx)</p><p></p><p>Taking x common from both numerator and denominator,</p><p></p><p>\sf v+x\:\dfrac{dv}{dx} =\dfrac{vx\:(1+v)}{x\:(1-v)}v+xdxdv=x(1−v)vx(1+v)</p><p></p><p>\sf v+x\:\dfrac{dv}{dx} =\dfrac{v\:(1+v)}{(1-v)}v+xdxdv=(1−v)v(1+v)</p><p></p><p>\sf v+x\:\dfrac{dv}{dx} =\dfrac{v+v^2}{(1-v)}v+xdxdv=(1−v)v+v2</p><p></p><p>\sf x\:\dfrac{dv}{dx} =\dfrac{v+v^2}{(1-v)}-vxdxdv=(1−v)v+v2−v</p><p></p><p>\sf x\:\dfrac{dv}{dx} =\dfrac{v+v^2-v+v^2}{(1-v)}xdxdv=(1−v)v+v2−v+v2</p><p></p><p>\sf x\:\dfrac{dv}{dx} =\dfrac{2v^2}{(1-v)}xdxdv=(1−v)2v2</p><p></p><p>Now making it variable separable,</p><p></p><p>\sf \dfrac{1-v}{2v^2} \: dv=\dfrac{1}{x}\: dx2v21−vdv=x1dx</p><p></p><p>Integrate on both sides,</p><p></p><p>\displaystyle \sf \int\limits {\dfrac{1-v}{2v^2} } \, dv =\int\limits {\dfrac{1}{x} } \, dx∫2v21−vdv=∫x1dx</p><p></p><p>\displaystyle \sf \int\limits {\dfrac{1}{2v^2} } \, dv -\int\limits {\dfrac{v}{2v^2} } \, dv =\int\limits {\dfrac{1}{x} } \, dx∫2v21dv−∫2v2vdv=∫x1dx</p><p></p><p>\displaystyle \sf \dfrac{1}{2} \int\limits {\dfrac{1}{v^2} } \, dv -\dfrac{1}{2} \int\limits {\dfrac{1}{v} } \, dv =\int\limits {\dfrac{1}{x} } \, dx21∫v21dv−21∫v1dv=∫x1dx</p><p></p><p>\sf -\dfrac{1}{2v} -\dfrac{1}{2}\: log|v|=log|x|+C−2v1−21log∣v∣=log∣x∣+C</p><p></p><p>Now we know that y = vx</p><p></p><p>Hence v = y/x</p><p></p><p>Substitute the value of v,</p><p></p><p>\sf -\dfrac{x}{2y} -\dfrac{1}{2}\: log|\dfrac{y}{x} |=log|x|+C−2yx−21log∣xy∣=log∣x∣+C</p><p></p><p>This is the solution of the given differential equation.</p><p></p><p>$