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.

bn,mn,n.vn.

step-by-step explanation:

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