Question

Find the integrating factor of the differential equation
(6xy-3y^2+2y) dx + 2(x-y) dy =0
and hence solve it

Answer 1

Answer on Question #66661 – Math – Differential Equations

Question

Find the integrating factor of the differential equation

$$(6xy - 3y^2 + 2y) \, dx + 2(x - y) \, dy = 0$$

and hence solve it.

Solution

$$P(x, y) \, dx + Q(x, y) \, dy = 0, \text{ where}$$ $$P(x, y) = 6xy - 3y^2 + 2y \text{ and } Q(x, y) = 2(x - y)$$

First, we need to check whether

$$\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}.$$ $$\frac{\partial P}{\partial y} = 6x - 6y + 2 \neq 2 = \frac{\partial Q}{\partial x};$$ $$\frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x} = 6x - 6y + 2 - 2 = 6x - 6y = 6(x - y)$$

It can be noticed that

$$\frac{1}{Q} \left( \frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x} \right) = \frac{6(x - y)}{2(x - y)} = 3$$

that means our integrating factor as a function depends on $x$ only:

$$\mu = \mu(x).$$

We can find it from the equation:

$$\frac{1}{ \mu} \frac{d\mu}{dx} = \frac{1}{Q} \left( \frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x} \right) = 3$$ $$\frac{d\mu}{\mu} = 3 \, dx$$ $$\int \frac{d\mu}{\mu} = 3 \int dx$$ $$\ln |\mu| = 3x + \ln |C|$$ $$\mu = C e^{3x},$$

but we will take the integrating factor

$$\mu = e^{3x}$$

Now, we have

$$\mathrm{e}^{3x}(6xy - 3y^2 + 2y)\mathrm{dx} + 2\mathrm{e}^{3x}(x - y)\mathrm{dy} = 0$$ $$P(x, y) = \mathrm{e}^{3x}(6xy - 3y^2 + 2y) \text{ and } Q(x, y) = 2\mathrm{e}^{3x}(x - y)$$ $$\frac{\partial P}{\partial y} = \mathrm{e}^{3x}(6x - 6y + 2) = \mathrm{e}^{3x}(-6y + 2 + 6x) = \frac{\partial Q}{\partial x}$$

So, $\exists u(x,y): \begin{cases} \dfrac{\partial u}{\partial x} = \mathrm{e}^{3x}(6xy - 3y^2 + 2y) \\ \dfrac{\partial u}{\partial y} = 2\mathrm{e}^{3x}(x - y) \end{cases}$

$$\begin{aligned} u(x, y) &= \int \mathrm{e}^{3x}(6xy - 3y^2 + 2y)\mathrm{dx} + \varphi(y) \\ &= \frac{1}{3} \int 6xyd(\mathrm{e}^{3x}) + \frac{1}{3}\mathrm{e}^{3x}(-3y^2 + 2y) + \varphi(y) = \\ &= \mathrm{e}^{3x}2xy - \frac{1}{3} \int \mathrm{e}^{3x}d(6xy) + \frac{1}{3}\mathrm{e}^{3x}(-3y^2 + 2y) + \varphi(y) = \\ &= \mathrm{e}^{3x}2xy - 2 \int \mathrm{e}^{3x}y\mathrm{dx} + \frac{1}{3}\mathrm{e}^{3x}(-3y^2 + 2y) + \varphi(y) = \\ &= \mathrm{e}^{3x}\left(2xy - \frac{2}{3}y - y^2 + \frac{2}{3}y\right) + \varphi(y) = \mathrm{e}^{3x}(2xy - y^2) + \varphi(y) \\ \end{aligned}$$ $$\begin{aligned} \frac{\partial u}{\partial y} &= \frac{\partial}{\partial y}\left(\mathrm{e}^{3x}(2xy - y^2) + \varphi(y)\right) = 2\mathrm{e}^{3x}(x - y) \\ \mathrm{e}^{3x}(2x - 2y) + \varphi'(y) &= 2\mathrm{e}^{3x}(x - y) \\ \varphi'(y) &= 0 \\ \varphi(y) &= C \\ u(x, y) &= \mathrm{e}^{3x}(2xy - y^2) + C = 0, \end{aligned}$$

where $C$ is an arbitrary real constant.

Answer: $u(x, y) = \mathrm{e}^{3x}(2xy - y^2) + C = 0$.

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