dx%2B(2y-sinx-x%5E3%2Bln-y)dy%3D0.jpeg "(y^2cosx-3x^2y-2x)dx+(2y Sinx-x^3+ln Y)dy=0")
Solving the Differential Equation: (y^2cosx - 3x^2y - 2x)dx + (2ysin x - x^3 + ln y)dy = 0
The given differential equation is:
(y^2cosx - 3x^2y - 2x)dx + (2ysin x - x^3 + ln y)dy = 0
This equation is a non-exact differential equation. We can check this by verifying if the following condition holds:
∂M/∂y = ∂N/∂x
Where:
- M = y^2cosx - 3x^2y - 2x
- N = 2ysin x - x^3 + ln y
Calculating the partial derivatives:
- ∂M/∂y = 2ycosx - 3x^2
- ∂N/∂x = 2ycosx - 3x^2
Since ∂M/∂y = ∂N/∂x, the equation is exact. This means we can find a potential function, u(x,y), such that:
- ∂u/∂x = M
- ∂u/∂y = N
To find u(x,y), we integrate M with respect to x:
u(x,y) = ∫(y^2cosx - 3x^2y - 2x)dx = y^2sinx - x^3y - x^2 + g(y)
Here, g(y) is an arbitrary function of y, as the partial derivative with respect to x will eliminate any function solely of y.
Now, we differentiate u(x,y) with respect to y and equate it to N:
∂u/∂y = 2ysinx - x^3 + g'(y) = 2ysin x - x^3 + ln y
This implies: g'(y) = ln y
Integrating both sides with respect to y: g(y) = yln y - y + C
Substituting g(y) back into the potential function:
u(x,y) = y^2sinx - x^3y - x^2 + yln y - y + C
The general solution to the differential equation is given by:
u(x,y) = C
Therefore, the solution to the given differential equation is:
y^2sinx - x^3y - x^2 + yln y - y = C
Where C is an arbitrary constant.