Question

2xy dy -(y²-x²) dx =0 verify the equation is homogeneous and solve ?​

Answer 1

Answer:

Solution of the differential equation = y² + x² = Cx

Step-by-step explanation:

Given:

The differential equation 2xy dy - (y² - x²) dx = 0

To Find:

To verify that the equation is homogeneous and to find the solution for it.

Solution:

The equation can be written as,

$\sf -(y^2-x^2)dx=-2xy\:dy$

$\sf y^2-x^2\: dx=2xy\: dy$

$\sf \dfrac{dy}{dx} =\dfrac{y^2-x^2}{2xy}---(1)$

Let,

$\sf F(x,y)=\dfrac{y^2-x^2}{2xy}$

Replacing x and y by λx and λy respectively,

$\sf F(\lambda x,\: \lambda y)=\dfrac{(\lambda y)^2-(\lambda x)^2}{2\times \lambda x\times \lambda y}$

$\sf \implies \dfrac{\lambda^2y^2-\lambda^2x^2}{2\times \lambda^2\times xy}$

$\sf \implies \dfrac{\lambda^2(y^2-x^2)}{\lambda^2(2xy)} =\lambda^0\:F(x,y)$

Therefore the given function is homogeneous and is of degree 0.

Finding the solution for it,

Put y = vx

Differentiate on both sides with respect to x and by using product rule,

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

Hence from equation 1,

$\sf v+x\: \dfrac{dv}{dx} =\dfrac{v^2x^2-x^2}{2\times x\times vx}$

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

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

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

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

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

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

Separating the variables,

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

Integrating on both sides,

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

Solving it we get,

$\sf log|v^2+1|=-log|x|+C$

$\sf log|v^2+1|+log|x|=C$

$\sf log|x(v^2+1)|=log\:C$

$\sf x(v^2+1)=C$

We know that y = vx, v = y/x

Hence,

$\sf x\times \bigg(\dfrac{y^2}{x^2} +1\bigg)=C$

$\sf \dfrac{y^2}{x} +x=C$

Multiplying the whole equation by x,

$\sf y^2+x^2=Cx$

Hence this is the solution of the given differential equation.