Question

What is the solution of the Homogeneous Differential Equation? : `dy/dx = (x^2+y^2-xy)/x^2` with `y(1)=0`

Answer 1

`y = (xln|x|)/(1+ln|x|)`

Explanation:

We have:

`dy/dx = (x^2+y^2-xy)/x^2` with `y(1)=0`

Which is a First Order Nonlinear Ordinary Differential Equation. Let us attempt a substitution of the form:

`y = vx`

Differentiating wrt `x` and applying the product rule, we get:

`dy/dx = v + x(dv)/dx`

Substituting into the initial ODE we get:

`v + x(dv)/dx = (x^2+(vx)^2-x(vx))/x^2`

Then assuming that `x ne 0` this simplifies to:

`v + x(dv)/dx = 1+v^2-v`

`:. x(dv)/dx = v^2-2v+1`

And we have reduced the initial ODE to a First Order Separable ODE, so we can collect terms and separate the variables to get:

`int \ 1/(v^2-2v+1) \ dv = int \ 1/x \ dx`

`int \ 1/(v-1)^2 \ dv = int \ 1/x \ dx`

Both integrals are standard, so we can integrate to get:

`-1/(v-1) = ln|x| + C`

Using the initial condition, `y(1)=0 => v(1)=0`, we get:

`-1/(0-1) = ln|1| + C => 1`

Thus we have:

`-1/(v-1) = ln|x| +1`

`:. 1-v = 1/(1+ln|x|)`

`:. v = 1 - 1/(1+ln|x|)`

`\ \ \ \ \ \ \= (1+ln|x|-1)/(1+ln|x|)`

`\ \ \ \ \ \ \= (ln|x|)/(1+ln|x|)`

Then, we restore the substitution, to get the General Solution:

`y/x = (ln|x|)/(1+ln|x|)`

`:. y = (xln|x|)/(1+ln|x|)`