Question

Solve the following differential equations:

1. 8(8-y²)^1/2 ln xdx + (8y-5) xdy = 0
2. dy/dx = ((x² - y²)/x²))^1/2 + y/x

Answer 1

Solution

1.

Use Method of seperation of variables.

$8(8-y^2)^{\frac12}ln(x)dx$ =-(8y-5)xdy

Rewrite as

$8(8-y^2)^{\frac12}lnxdx$ =(5-8y)xdy

By separation,

$\frac{8lnx}{x}dx=\frac{5-8y}{(8-y^2)^{\frac12}}dy$

Rewrite as

$\frac{8ln(x)}{x}dx=\frac{5}{\sqrt{8-y^2}}dy-8(\frac{y}{\sqrt{8-y^2}})dy$

Integrate both sides:

$4ln^2(x)+C=5arcsin(\frac{y}{2\sqrt2})+8\sqrt{8-y^2}$

2)

Equation is homogeneous.

Take y=vx

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

By substitution,

$v+x\frac{dv}{dx}=\sqrt{1-v^2}+v$

By simplication,

$x\frac{dv}{dx}=\sqrt{1-v^2}$

Separate by variables:

$\frac{1}{\sqrt{1-v^2}}dv=\frac1xdx$

Integrate both sides:

arcsin(v)=ln(x)+C

But $v=\frac yx$ ,Replace back:

$arcsin(\frac yx)=ln(x)+C$