Question

Solve the differential equation:
(x y2 -x2)dx + (3x2y2+x2y-2x3+y2)dy=0 .

Answer 1

Answer on Question #40385, Math, Differential Calculus

Task was solve the differential equation:

$$(x y ^ {2} - x ^ {2}) d x + \left(3 x ^ {2} y ^ {2} + x ^ {2} y - 2 x ^ {3} + y ^ {2}\right) d y = 0$$

Solution:

Let's find the integrating factor $\mu(y)$ by which you multiply the equation, that it was resolved in the total derivatives. For this let $R(x,y) = xy^2 - x^2$ , $S(x,y) = 3x^2y^2 + x^2y - 2x^3 + y^2$ .

Now we have equation like $\mu (y)R(x.y) + \mu (y)S(x,y)\frac{dy}{dx} = 0$

Condition for the existence the integrating factor $\mu (y)$ is $\frac{\partial}{\partial y}\bigl (\mu (y)R(x.y)\bigr) = \frac{\partial}{\partial}\bigl (\mu (y)S(x,y)\bigr).$

Substitute $R(x,y)$ and $S(x,y)$ into (1):

$$\frac {d \mu (y)}{d y} \left(x y ^ {2} - x ^ {2}\right) + 2 y x \mu (y) = \mu (y) \left(6 x y ^ {2} + 2 x y - 6 x ^ {2}\right)$$

Collect the common multipliers

$$\frac {d \mu (y)}{d y} \left(x y ^ {2} - x ^ {2}\right) = \mu (y) \left(6 x y ^ {2} - 6 x ^ {2}\right)$$ $$\frac {d \mu (y)}{\mu (y)} = \frac {\left(6 x y ^ {2} - 6 x ^ {2}\right)}{\left(x y ^ {2} - x ^ {2}\right)} d y = 6 d y.$$

Integrate to $y$ : In $\mu (y) = 6y = > \mu (y) = e^{6y}$ .

Now we found integrating factor $\mu(y)$ . Multiply our equation for $\mu(y)$ :

$$e ^ {6 y} \left(x y ^ {2} - x ^ {2}\right) d x + e ^ {6 y} \left(3 x ^ {2} y ^ {2} + x ^ {2} y - 2 x ^ {3} + y ^ {2}\right) d y = 0$$

Let $P(x,y) = e^{6y}(xy^2 - x^2), Q(x,y) = e^{6y}(3x^2y^2 + x^2y - 2x^3 + y^2)$

This is an exact equation, because $\frac{\partial P(x,y)}{\partial y} = 6xe^{6y}(y^2 - x) + 2xe^{6y}y = \frac{\partial Q(x,y)}{\partial x}$ .

Define $f(x, y)$ such that $\frac{\partial f(x, y)}{\partial x} = P(x, y), \frac{\partial f(x, y)}{\partial y} = Q(x, y)$ .

Then solution will be given by $f(x,y) = C$ , where $C$ is arbitrary constant.

Integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to $x$ in order to find $f(x,y)$ :

$f(x,y) = \int e^{6y}(xy^2 -x^2)dx = e^{6y}\left(\frac{y^2x^2}{2} -\frac{x^3}{3}\right) + g(y)$ , where $g(y)$ is an arbitrary function of $y$ .

Differentiate $f(x, y)$ with respect to $y$ in order to find $g(y)$ :

$$\frac {\partial f (x , y)}{\partial y} = \frac {\partial}{\partial y} \left(e ^ {6 y} \left(\frac {y ^ {2} x ^ {2}}{2} - \frac {x ^ {3}}{3}\right) + g (y)\right) = e ^ {6 y} y x ^ {2} + 6 e ^ {6 y} \left(\frac {y ^ {2} x ^ {2}}{2} - \frac {x ^ {3}}{3}\right) + \frac {d g (y)}{d y}$$

Substitute into $\frac{\partial f(x,y)}{\partial y} = Q(x,y)$ :

$$e ^ {6 y} y x ^ {2} + 6 e ^ {6 y} \left(\frac {y ^ {2} x ^ {2}}{2} - \frac {x ^ {3}}{3}\right) + \frac {d g (y)}{d y} = e ^ {6 y} \left(3 x ^ {2} y ^ {2} + x ^ {2} y - 2 x ^ {3} + y ^ {2}\right)$$ $$e ^ {6 y} y x ^ {2} + e ^ {6 y} \left(3 y ^ {2} x ^ {2} - 2 x ^ {3}\right) + \frac {d g (y)}{d y} = e ^ {6 y} \left(3 x ^ {2} y ^ {2} + x ^ {2} y - 2 x ^ {3} + y ^ {2}\right)$$ $$\frac {d g (y)}{d y} = e ^ {6 y} y ^ {2}$$

Integrate $\frac{dg(y)}{dy}$ with respect to $y$:

$$\begin{array}{l} g(y) = \int e^{6y} y^2 dy = \frac{1}{6} \int y^2 de^{6y} = \frac{1}{6} y^2 e^{6y} - \frac{1}{6} \int e^{6y} \cdot 2 y dy = \frac{1}{6} y^2 e^{6y} - \frac{1}{3} \left(\frac{1}{6} y e^{6y} - \frac{1}{6} \int e^{6y} dy\right) = \\ = \frac{1}{6} y^2 e^{6y} - \frac{1}{3} \left(\frac{1}{6} y e^{6y} - \frac{1}{36} e^{6y}\right) = \frac{1}{108} e^{6y} (18y^2 - 6y + 1) \end{array}$$

Substitute $g(y)$ into $f(x,y)$:

$$f(x, y) = \int e^{6y} (xy^2 - x^2) dx = e^{6y} \left(\frac{y^2 x^2}{2} - \frac{x^3}{3}\right) + \frac{1}{108} e^{6y} (18y^2 - 6y + 1)$$

The solution is $f(x,y) = C$

$$e^{6y} \left(\frac{y^2 x^2}{2} - \frac{x^3}{3}\right) + \frac{1}{108} e^{6y} (18y^2 - 6y + 1) = C$$

Answer:

$$e^{6y} \left(\frac{y^2 x^2}{2} - \frac{x^3}{3}\right) + \frac{1}{108} e^{6y} (18y^2 - 6y + 1) = C$$