Question

Solve (x²y-2xy²)dx-(x³-3x²y) dy =0

Answer 1

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Answer 2

Concept:

We need to first recall the concept of homogeneous differential equation to solve this question.

• A differential equation (D.E.)  of the form

       f(x,y) =  $x^{n}F(\frac{y}{x} )$

       is called homogeneous equation of degree n.

• A Solution of homogeneous D.E. is obtained by substituting y = vx and then convert it into separable form and then integrating.

Given:

The differential equation of the form:

(x²y - 2xy²) dx - (x³ - 3x²y) dy =0

To find:

The solution of the given differential equation.

Solution:

The given differential equation can be written as:

$\frac{dy}{dx} = \frac{x^{2}y-2xy^{2}}{x^{3}-3x^{2}y}$

$\frac{dy}{dx} = \frac{y(x-2y)}{x(x-3y)}$           (1)

It is a homogeneous differential equation of degree 1.

On substituting y = vx

$\frac{dy}{dx} = v+x\frac{dv}{dx}$    

so,   equation (1) becomes:

 $v+x\frac{dv}{dx}=v(\frac{1-2v}{1-3v})$    

       $x\frac{dv}{dx}=(\frac{v-2v^{2} }{1-3v})-v$

       $x\frac{dv}{dx}=(\frac{v-2v^{2}-v+3v^{2} }{1-3v})$

       $x\frac{dv}{dx}=(\frac{v^{2} }{1-3v})$

$\frac{(1-3v)}{v^{2}} dv= \frac{1}{x} dx$

On integrating ,

$\int \frac{1}{v^{2} } \, dv - \int \frac{3}{v} \, dv =\int \frac{1}{x } \, dx$

      $-\frac{1}{v} - 3 log v = log x + log C$

                   $-\frac{1}{v} = 3 log v + log x + log C$

                   $-\frac{1}{v}=log cx v^{3}$

Now substituting v= y/x we get,

                  $-\frac{x}{y}=log \frac{ cy^{3}}{x^{2} }$

                  $cy^{3}=x^{2} e^{-\frac{x}{y}$

Hence, the solution of given D.E. is given by:  $y^{3}=cx^{2} e^{-\frac{x}{y}$.