dx-(x%5E2-3xy)dy%3D0.jpeg "(xy-2y^2)dx-(x^2-3xy)dy=0")
Solving the Differential Equation: (xy - 2y^2)dx - (x^2 - 3xy)dy = 0
This article will guide you through the process of solving the given differential equation:
(xy - 2y^2)dx - (x^2 - 3xy)dy = 0
This equation is a homogeneous differential equation because we can rewrite it in the form:
M(x, y)dx + N(x, y)dy = 0
where M(x, y) and N(x, y) are homogeneous functions of the same degree. In our case, both M(x, y) = xy - 2y^2 and N(x, y) = -x^2 + 3xy are homogeneous functions of degree 2.
Here's how to solve the equation:
1. Substitution and Simplification
We can solve homogeneous differential equations using the substitution y = vx. This gives us:
- dy = vdx + xdv
Substituting these into our original equation, we get:
(x(vx) - 2(vx)^2)dx - (x^2 - 3x(vx))(vdx + xdv) = 0
Simplifying the equation:
(vx^2 - 2v^2x^2)dx - (x^2 - 3vx^2)(vdx + xdv) = 0
2. Separating Variables
Now, we can separate the variables 'x' and 'v':
(vx^2 - 2v^2x^2 - x^2v + 3v^2x^2)dx = (x^2 - 3vx^2)xdv
Combining like terms and factoring:
x^2(v + 2v^2)dx = x^3(1 - 3v)dv
Dividing both sides by x^2(1 - 3v)(v + 2v^2):
dx / x = (1 - 3v) / (v + 2v^2) dv
3. Integrating Both Sides
Now we can integrate both sides of the equation:
∫(1/x)dx = ∫((1 - 3v) / (v + 2v^2))dv
The left side integrates to ln|x|. For the right side, we need to use partial fractions.
4. Partial Fractions
Decomposing the right side:
(1 - 3v) / (v + 2v^2) = A / v + B / (1 + 2v)
Solving for A and B, we get:
- A = 1
- B = -4
Now the integral becomes:
∫(1/x)dx = ∫(1/v)dv - ∫(4 / (1 + 2v))dv
5. Solving the Integral
Integrating both sides:
ln|x| = ln|v| - 2ln|1 + 2v| + C
where C is the constant of integration.
6. Back Substitution and Solution
Substituting back v = y/x, we get:
ln|x| = ln|y/x| - 2ln|1 + 2(y/x)| + C
Simplifying:
ln|x| = ln|y| - ln|x| - 2ln|x + 2y| + C
Combining logarithms and exponentiating both sides:
x^2 = Ce^(y/(x + 2y))
This is the general solution to the given differential equation.
7. Final Notes
- Remember that the constant of integration 'C' can be any real number.
- We can find a particular solution for a given initial condition.
- The solution might not be defined for all values of x and y.
This article provides a comprehensive guide to solving the given differential equation. However, always remember to double-check your calculations and use appropriate methods for each step.