Answer to Question #88345 in Differential Equations for Shivam Nishad

Question

Solve the following differential equations:
i) (x²+y²)dy/dx=xy

Answer 1

Solution. We write the equation as

$$(x^2+y^2)dy=xydx$$

$$-xydx+(x^2+y^2)dy=0$$

Consider the equation as

$$M(x,y)dx+N(x,y)dy=0.$$

Find

$$\frac {\delta M} {\delta y} = - x$$

$$\frac {\delta N} {\delta x} = 2x$$ $$\frac {\delta M} {\delta y} =\not \frac {\delta N} {\delta x}$$

Find the value

$$-\frac {\frac {\delta M} {\delta y}-\frac {\delta N} {\delta x}} {M(x,y)} =-\frac {-x-2x} {-xy} =- \frac {3} {y}.$$

The resulting function depends only on y. Therefore, find the integrating factor using the equation

$$\frac {k'(y)} {k(y)} = -\frac {3} {y}.$$

Hence

$$k(y)=\frac {1} {y^3}$$

.

Get the total (exact) differential equation

$$-\frac {x} {y^2}dx+(\frac {x^2} {y^3}+\frac {1} {y})dy=0.$$

Let a solution to the equation be U(x,y). Hence

$$\frac {\delta U} {\delta x} = -\frac {x} {y^2}$$

Get

$$U=- \frac {x^2} {2y^2} +C(y)$$

where C(y) is function of y. Therefore

$$\frac {\delta U} {\delta y} =\frac {x^2} {y^3}+C'(y)=\frac {x^2} {y^3}+\frac {1} {y}$$

$$C'(y)=\frac {1} {y}$$ $$С(y)=ln(y) + C$$

where C is constant. Hence

$$U(x,y)=- \frac {x^2} {2y^2}+ln(y)+C=k,$$

where k is constant.

Answer: 

$$U(x,y)=- \frac {x^2} {2y^2}+ln(y)=D$$

where D is constant.