\(y=vx\)
\(v=yx\)
\(x=vy\)
\(x=v\)
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.
A differential equation having the formation f(x,y)dy = g(x,y)dx is known to be homogeneous differential equation if the degree of f(x,y) and g(x, y) is entirely same. A function of form F(x,y), written in the formation of kn F(x,y) is called a homogeneous function of degree n, for k≠0. Therefore, f and g are the homogeneous functions of the same degree of x and y. Here, the change of variable y = ux directs to an equation of the form;
dx/x = h(u) du which could be easily desegregated.
To solve a homogeneous differential equation go through the following steps:-
Given the differential equation of the type