Solve the Differential Equation using differential formulae

ordinary differential equations

In these questions, we are allowed to use the following differential formulae (or anything similar or combination of these):
$$
\begin{split}
d\left(\frac{x}{y}\right) &= \frac{ydx-xdy}{y^2} \\
\\
d(xy) &= ydx+xdy\\
\\
d\left(\ln\left(\frac{x}{y}\right)\right) &= \frac{ydx-xdy}{xy}\\
\\
(x^2+y^2) &= 2(xdx+ydy) \\
\\
d\left(\tan^{-1}\left(\frac{x}{y}\right)\right)&=\frac{ydx-xdy}{x^2+y^2}
\end{split}$$

The three problems I am stuck at are:

  1. $$(y-xy^2)dx+(x+x^2y^2)dy =0 $$
    Attempt
    $$ydx+xdy = xy^2dx-x^2y^2dy$$
    I noticed the LHS could be multiple things such as $d\left(\ln\left(\frac{x}{y}\right)\right)\cdot xy$, $d(xy)$ etc…
    Then, I tried this:
    $$
    \begin{split}
    xyd(\ln(xy) &=xy^2(dx-xdy)\\
    \implies d(\ln(xy))&=y(dx-xdy)
    \end{split}$$
    Now this is where I am stuck, I have to express the RHS either in some differential formula (which I am unable to do) or to re-express the LHS so that after simplifying, the RHS will only be of "1 variable".

  2. $$x dy -ydx = x^2y^4(xdy+ydx)$$

  3. $$xdy+ydx+x^2y^5dy=0$$
    I tried using combination of the above formulae in all of them, such as $d\left(\ln\right(\frac{x^2}{y^2}\left)\right) $ etc., but couldn't solve it this way.

I got the solutions using integrating factors and exact equations, as follows:

Solution of 1. $$-\frac{1}{xy}-\ln(x)+y=c$$

Solution of 2. $$3x+x^3y^4+cy=0$$

Solution of 3. $$x(y^5+cy)=4$$

Thanks in advance.

Best Answer

I solved them.

I will post the answers so that if it is asked in the future, then the steps are provided here.

For (i): $$y dx + xdy = xy^2dx-x^2y^2dy$$ $$\implies d(xy) = x^2y^2(\frac{dx}{x}-dy)$$ $$\implies \frac{d(xy)}{x^2y^2}=d(\ln(x)-y) $$ $$\implies \int \frac{d(xy)}{(xy)^2} = \int d(\ln(x)-y) $$ $$\implies -\frac{1}{xy} = \ln(x)-y + c \iff -\frac{1}{xy}-\ln(x)+y=c$$

For (ii): $$xdy-ydx =x^2y^4(xdy+ydx)$$ $$\implies x^2d(\frac{y}{x})=x^2y^4d(xy) $$ $$\implies d(\frac{y}{x})=y^4d(xy)$$

Here we multiply both sides by $\frac{x^2}{y^2}:$ $$\implies (\frac{y}{x})^{-2}d(\frac{y}{x})=(xy)^2d(xy) $$ $$\implies \int (\frac{y}{x})^{-2}d(\frac{y}{x})= \int (xy)^2d(xy) $$ $$\implies -\frac{x}{y}=\frac{(xy)^3}{3}+c $$ $$\implies -3x=x^3y^4+cy \iff 3x+x^3y^4+cy=0 $$

For (iii): $$xdy+ydx+x^2y^5=9$$ $$\implies d(xy)=-(xy)^2y^3dy $$ $$\implies -\frac{d(xy)}{(xy)^2} = y^3dy$$ $$\implies \int -\frac{d(xy)}{(xy)^2}=\int y^3 dy $$ $$\implies \frac{1}{xy}=\frac{y^4}{4}+c \iff x(y^5+cy)=4 $$

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