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

1 Answer
May 17, 2018

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|)