Answer to Question #237563 in Differential Equations for Cha

Question

For equation of y^2dy= x(xdy-ydx)e^(x/y)

1. Find the function M(x,y) and its degree
2. Find the function N(x,y) and its degree
3. Is the given Differential Equation a homogenous? If yes, what is the degree?

Answer 1

Solution;

Rewrite the equation as follows;

$y^2dy=(x^2dy-xydx)e^{\frac xy}$

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

Combine as follows;

$xye^{\frac xy}dx=(x^2e^{\frac xy}-y^2)dy$ ...(a)

1.)

From the above equation (a),

$M(x,y)=xye^{\frac xy}dx$

The degree of M(x,y) is 1.

2.)

From the above equation (a);

$N(x,y)=x^2e^{\frac xy}-y^2$

The degree of N(x,y) is 2.

3.)

Because of $e^{\frac xy}$ ,take $\frac{dx}{dy}$ instead of $\frac{dy}{dx}$

From equation (a);

We obtain;

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

Now put ,F(x,y)=$\frac{dx}{dy}$ and find $F(\lambda x,\lambda y)$

Hence we have ;

$F(\lambda x,\lambda y)=\frac {(\lambda x)^2e^{\frac{\lambda x}{\lambda y}}-(\lambda y)^2}{(\lambda x)(\lambda y)e^{\frac{\lambda x}{\lambda y}}}$

$F(\lambda x,\lambda y)=\frac{(\lambda )^2[x^2e^{\frac xy}-y^2]}{\lambda^2xye^{\frac xy}}$ =F(x,y)

Clearly;

$F(\lambda x,\lambda y)=F(x,y)=\lambda ^0F(x,y)$

Therefore ,the given equation is a homogeneous function of degree zero(0).