What is the solution of the Homogeneous Differential Equation? : dy/dx=(x^2+y2^−xy)/(x^2) with y(1)=0

allbleachix 2022-09-12 Answered
What is the solution of the Homogeneous Differential Equation? : d y d x = x 2 + y 2 - x y x 2 with y(1)=0
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Answers (1)

Koen Henson
Answered 2022-09-13 Author has 17 answers
This is a First Order Nonlinear Ordinary Differential Equation. Let us attempt a substitution of the form:
y = v x
Differentiating wrt x and applying the product rule, we get:
d y d x = v + x d v d x
Substituting into the initial ODE we get:
v + x d v d x = x 2 + ( v x ) 2 - x ( v x ) x 2
Then assuming that x 0 this simplifies to:
v + x d v d x = 1 + v 2 - v
x d v d x = v 2 - 2 v + 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:
  1 v 2 - 2 v + 1   d v =   1 x   d x
  1 ( v - 1 ) 2   d v =   1 x   d x
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 = x ln | x | 1 + ln | x |

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