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Find the integrating factor for the equation: $(3x+\frac{6}{y})\mathrm{d}x+(\frac{x^2}{y}+\frac{3y}{x})\mathrm{d}y=0$.

Write $P_1(x,y)=2x+\frac{6}{y}$, $Q_1(x,y)=\frac{x^2}{y}$, $P_2(x,y)=x$, and $Q_2(x,y)=\frac{3y}{x}$, then $P_1(x,y)\mathrm{d}x+Q_1(x,y)\mathrm{d}y+P_2(x,y)\mathrm{d}x+Q_2(x,y)\mathrm{d}y=0$.

For $P_1(x,y)\mathrm{d}x+Q_1(x,y)\mathrm{d}y$, notice that $\frac{1}{P_1(x,y)}(\frac{\partial Q_1(x,y)}{\partial x}-\frac{\partial P_1(x,y)}{\partial y})=\frac{1}{y}$, and the integrating factor for this part is $\mu_1(y)=\mathrm{e}^{\int\frac{1}{y}\mathrm{d}y}=y$. Hence, by computing $\int_{x_0}^x\mu_1(y)P_1(x,y)\mathrm{d}x+\int_{y_0}^y\mu_1(y)Q_1(x_0,y)\mathrm{d}y$, one obtains $\Phi_1(x,y)=x^2y+6x$.

Similary, for $P_2(x,y)\mathrm{d}x+Q_2(x,y)\mathrm{d}y$, notice that $\frac{1}{Q_2(x,y)}(\frac{\partial P_2(x,y)}{\partial y}-\frac{\partial Q_2(x,y)}{\partial x})=\frac{1}{x}$, and the integrating factor for this part is $\mu_2(x)=\mathrm{e}^{\int\frac{1}{x}\mathrm{d}x}=x$. Hence, by computing $\int_{x_0}^x\mu_2(y)P_2(x,y)\mathrm{d}x+\int_{y_0}^y\mu_2(y)Q_2(x_0,y)\mathrm{d}y$, one obtains $\Phi_2(x,y)=\frac{x^3}{3}+\frac{3y^2}{2}$.

In order to find the integrating factor for the whole equation, one needs to find two smooth functions $g_1(t)$ and $g_2(t)$ such that $\mu_1g_1(\Phi_1(x,y))=\mu_2g_2(\Phi_2(x,y))$, i.e., $yg_1(x^2y+6x)=xg_2(\frac{x^3}{3}+\frac{3y^2}{2})$.

However, it seems impossible, does not it?

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$$(3x+\frac{6}{y})\mathrm{d}x+(\frac{x^2}{y}+\frac{3y}{x})\mathrm{d}y=0 $$ Simplification (reduction to common denominator) : $$(3x^2y+6x)\mathrm{d}x+(x^3+3y^2)\mathrm{d}y=P(x,y)\mathrm{d}x+Q(x,y)\mathrm{d}y=0 \tag 1$$ $$\begin{cases}\frac{\partial P}{\partial y}=3x^2\\ \frac{\partial Q}{\partial x}=3x^2 \end{cases} \quad\to\quad \frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$$ Thus equation $(1)$ is a total differential.

So, there is no need for integrating factor (Or the integrating factor $=1$ ).

This is equivalent to say that $xy$ is an integrating factor for the initial equation, before simplification. $$dF=(3x^2y+6x)\mathrm{d}x+(x^3+3y^2)\mathrm{d}y=0 \tag 2$$

$$\begin{cases} \frac{\partial F}{\partial x}=3x^2y+6x \quad\to\quad F=\int(3x^2y+6x)dx=x^3y+3x^2+C_1(y)\\ \frac{\partial F}{\partial y}=x^3+3y^2 \quad\to\quad F=\int(x^3+3y^2)dy=x^3y+y^3+C_2(x) \end{cases}$$ $$F=x^3y+3x^2+C_1(y)=x^3y+y^3+C_2(x) \quad\to\quad \begin{cases} C_1(y)=y^3\\ C_2(x)=3x^2 \end{cases} \quad\to\quad F=x^3y+3x^2+y^3$$ From $(2) \quad dF=0\quad : \quad F=$constant. The solution of the ODE, expressed on the form of implicit equation is : $$x^3y+3x^2+y^3=C$$

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The integrating factor is $xy$.

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

Because upon clearing the fractions we find the resulting equation is exact.

\begin{eqnarray} F(x,y)&=&x^3y+3x^2+c_1(y)\\ &=&x^3y+y^3+c_2(x) \end{eqnarray}

Therefore

$$ x^3y+3x^2+y^3=c$$

P.S. As a teacher I have always found these "trick" questions to be somewhat unfair. After learning a complicated method for performing some operation, the textbook slips in a question which has a simple answer if you DON'T use the complex method just learned.

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  • $\begingroup$ Thank you for your answer. Yes, I agree with you. In fact, even in many cases that $P(x,y)$ and $Q(x,y)$ are polynomials, there is no natural way to find the integrating factor, especially in the step to find the smooth functions $g_1(t)$ and $g_2(t)$. $\endgroup$ – Hebe Jul 27 '17 at 7:39

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