# 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 # 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

# 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|) #