solve
(4x^3y^3+1/x)dx+(3x^4y^2-1/y)dy=0
(4x^3y^3+1/x)dx+(3x^4y^2-1/y)dy=0
To solve the given equation:
(4x^3y^3 + 1/x)dx + (3x^4y^2 - 1/y)dy = 0
We will use the method of exact differential equations. For a given equation (Mdx + Ndy = 0) to be exact, the following condition must be satisfied:
∂M/∂y = ∂N/∂x
Let's calculate the partial derivatives:
∂/∂y(4x^3y^3 + 1/x) = 12x^3y^2
∂/∂x(3x^4y^2 - 1/y) = 12x^3y^2
After evaluating the partial derivatives, we find that ∂M/∂y = ∂N/∂x.
Therefore, the given differential equation is exact.
To find the solution, we need to integrate M with respect to x and treat y as a constant, and then integrate N with respect to y and treat x as a constant. The constants of integration will be determined later.
Integrating Mdx with respect to x:
∫(4x^3y^3 + 1/x)dx = ∫4x^3y^3 dx + ∫1/x dx
= (4/4)x^4y^3 + ln|x| + C1
= x^4y^3 + ln|x| + C1
Here, C1 is the constant of integration.
Integrating Ndy with respect to y:
∫(3x^4y^2 - 1/y)dy = ∫3x^4y^2 dy - ∫1/y dy
= (3/3)x^4y^3 - ln|y| + C2
= x^4y^3 - ln|y| + C2
Here, C2 is the constant of integration.
The general solution is given by equating the integrated expressions of M and N:
x^4y^3 + ln|x| + C1 = x^4y^3 - ln|y| + C2
We can simplify this equation by combining the constants of integration:
ln|y| - ln|x| = C1 - C2
ln|y/x| = C
Here, C is the new constant obtained from the combination of C1 and C2.
Finally, we can express y in terms of x and C:
y = Cx
Therefore, the solution to the differential equation is y = Cx, where C is a constant.