solve homogeneous differential equation subject to y (1) = 1 by using appropriate substitution. dy/dx=y/x+x/y-1

A Differential Equation is an equation with a function and one or more of its derivatives:

differential equation dy/dx = 5xy

Example: an equation with the function y and its derivative dydx

Here we look at a special method for solving "Homogeneous Differential Equations"

Homogeneous Differential Equations

A first order Differential Equation is Homogeneous when it can be in this form:

dydx = F (yx)

We can solve it using Separation of Variables but first we create a new variable v = yx

v = yx which is also y = vx

And dydx = d (vx) dx = v dxdx + x dvdx (by the Product Rule)

Which can be simplified to dydx = v + x dvdx

Using y = vx and dydx = v + x dvdx we can solve the Differential Equation.

An example will show how it is all done:

Example: Solve dydx = x2 + y2xy

Can we get it in F (yx) style?

Start with: x2 + y2xy

Separate terms: x2xy + y2xy

Simplify: xy + yx

Reciprocal of first term: (yx) - 1 + yx

Yes, we have a function of (y/x).

So let's go:

Start with: dydx = (yx) - 1 + yx

y = vx and dydx = v + x dvdx : v + x dvdx = v-1 + v

Subtract v from both sides: x dvdx = v-1

Now use Separation of Variables:

Separate the variables: v dv = 1x dx

Put the integral sign in front: ∫v dv = ∫ 1x dx

Integrate: v22 = ln (x) + C

Then we make C = ln (k) : v22 = ln (x) + ln (k)

Combine ln: v22 = ln (kx)

Simplify: v = ±√ (2 ln (kx))

Now substitute back v = yx

Substitute v = yx: yx = ±√ (2 ln (kx))

Simplify: y = ±x √ (2 ln (kx))

And we have the solution.