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|) #

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|) #
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Answer 2

The solution to the homogeneous differential equation dy/dx = (x^2 + y^2 - xy)/x^2 with the initial condition y(1) = 0 is y(x) = x * (1 - √(2 - x^2)).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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