Answer to Question #159528 in Differential Equations for Ojugbele Daniel

Question

Solve the differential equations using homogeneous equation.

1. ( 2y - x )dy/dx = 2x + y given that y=3 and x=2.

2. ( xy + y square) + ( x square - xy) dy/dx =0

Answer 1

1. (2y - x )y' = 2x + y

We use the substitution: y = xu, y' = u+xu', then

(2xu-x)(u+xu') = 2x+xu

$x\frac{du}{dx}= \frac{2+u}{2u-1}-u = \frac{2+u - 2u^2+u}{2u-1} = -2\frac{u^2-u-1}{2u-1}$

This is an equation with separable variables.

Consider one of the possible solutions - with constant functions u(x) such that u²-u-1=0

Then $u(x) = (1\pm\sqrt{5})/2$ and $y(x) = x(1\pm\sqrt{5})/2$.

These solutions do not satisfy to the initial condition y(2) = 3.

$\frac{2u-1}{u^2-u-1}du = -\frac{2}{x}dx=-2d(\ln |x|)=-d(\ln x^2)$

$\frac{2u-1}{u^2-u-1}du = \frac{d(u^2-u)}{u^2-u-1} = d(\ln|u^2-u-1|)$ , therefore,

$d(\ln(x^2|u^2-u-1|))=0$

$d(x^2(u^2-u-1))=0$

$x^2(u^2-u-1)=c$

$u^2-u-1=cx^{-2}$

$y^2-xy-x^2=c$

$c=y(2)^2-2y(2)-2^2=9-6-4=-1$

$y^2-xy-x^2+1=0$

$y=(x\pm\sqrt{5x^2-4})/2$

$3 = y(2) = (2\pm 4)/2$ , therefore, one should choose the sign "+".

Answer. $y=(x+\sqrt{5x^2-4})/2$

2. $( xy + y^2) + ( x^2 - xy) dy/dx =0$

We use the substitution: y = xu, y' = u+xu', then

$x^2( u + u^2) + x^2 (1 - u) (u+xu') =0$

$2u+x(1-u)u' = 0$

$u=0$ or $\frac{u-1}{2u}du = \frac{dx}{x}$

$y=0$ or $u/2 -\ln|u|=\ln|x|+c$

$y=0$ or $y/(2x) -\ln|y| = c$

$y=0$ or $y -x\ln y^2 = 2cx$

Answer. $y=0$ or $y -x\ln y^2 = 2cx$