Question

homogeneous equation;
dy/dx=[2(2y-x)]/x+y; initial value of y(1)=2

Answer 1

\[\large{\frac{dy}{dx}=\frac{\frac{4y}{x}-\frac{2x}{x}}{\frac{x}{x}+\frac{y}{x}}}\]
=\[\frac{dy}{dx}=\frac{4 \frac{y}{x} -2 }{1+\frac{y}{x}}\]
let u=y/x
y=ux
y'=u+u'x

\[u+u'x=\frac{4u-2}{1+u}\]\[u'x=\frac{4u-2-u(1+u)}{1+u}\]

Answer 2

\[u'x=\frac{4u-2-u-u^2}{1+u} = \frac{-u^2+3u-2}{1+u}\]

Answer 3

ı found c=0 what should I do after?

Answer 4

now its seprable , after u integrate and find u ,
you substitute u=y/x in the equation and solve for y

Answer 5

c? how did u get that?

Answer 6

u'=du/dx

Answer 7

separable and we have to integrate

Answer 8

yeah did u integrate it?

Answer 9

yes and after ( u-2)^3/(u-1)^2=c/x after u=y/x y=2 and x=1 therefore c=0