Question

(x^2+y^2+2x)dx+2xydy=0​

Answer 1

$\displaystyle(x^2+y^2+2x)dx+2xydy=0\\\begin{cases}\displaystyle\frac{\delta F}{\delta x}=x^2+y^2+2x\\\displaystyle\frac{\delta F}{\delta y}=2xy\end{cases}\\F=\int 2xydy=xy^2+\phi(x)\\\frac{\delta F}{\delta x}=y^2+\phi'(x)=x^2+y^2+2x\\\phi'(x)=x^2+2x\\\phi(x)=\int(x^2+2x)dx=\frac{x^3}{3}+x^2+C\\F=\frac{x^3}{3}+x^2+xy^2+C$

Answer 2

The given equation is a first-order ordinary differential equation that is not separable. However, it appears to be exact, which means there exists a function of two variables whose partial derivatives satisfy the equation. To solve this type of equation, we need to check if it is exact by verifying whether the mixed partial derivatives are equal:

∂(x^2 + y^2 + 2x)/∂y = 2y ∂(2xy)/∂x = 2y

Since the mixed partial derivatives are equal, the equation is exact. To find the solution, we need to find a function Φ(x, y) such that its partial derivatives match the coefficients of the given equation:

∂Φ/∂x = x^2 + y^2 + 2x ∂Φ/∂y = 2xy

Let's find Φ(x, y) by integrating the first equation with respect to x and the second equation with respect to y:

∫(x^2 + y^2 + 2x)dx = ∫∂Φ/∂x dx ∫2xy dy = ∫∂Φ/∂y dy

Integrating the first equation with respect to x: ∫(x^2 + y^2 + 2x)dx = (1/3)x^3 + xy^2 + x^2 + C(y)

Here, C(y) is an arbitrary function of y.

Integrating the second equation with respect to y: ∫2xy dy = x*y^2 + D(x)

Here, D(x) is an arbitrary function of x.

Now, we equate the results of the two integrations to find Φ(x, y):

(1/3)x^3 + xy^2 + x^2 + C(y) = x*y^2 + D(x)

To simplify this, we can compare the terms involving x and y separately:

(1/3)x^3 + x^2 + C(y) = D(x) xy^2 = x*y^2

It's clear that the second equation is satisfied. So, we can equate the coefficients of the terms involving x:

(1/3)x^3 + x^2 = D(x)

To find D(x), integrate the left side with respect to x:

D(x) = ∫((1/3)x^3 + x^2)dx = (1/12)x^4 + (1/3)x^3 + K

Here, K is an integration constant.

So, the potential function Φ(x, y) is given by:

Φ(x, y) = (1/3)x^3 + xy^2 + (1/12)x^4 + (1/3)x^3 + K(y)

The solution to the given differential equation is Φ(x, y) = constant.