What is the solution of the Homogeneous Differential Equation? : dy/dx = (x^2+y^2-xy)/x^2 with y(1)=0
1 Answer
y = (xln|x|)/(1+ln|x|)
Explanation:
We have:
dy/dx = (x^2+y^2-xy)/x^2 withy(1)=0
Which is a First Order Nonlinear Ordinary Differential Equation. Let us attempt a substitution of the form:
y = vx
Differentiating wrt
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
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,
-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|)