Differential equation: $\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}$

Question

The solution of $\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}$ is given by:

a) $(x+2)^4 (1+\frac{2y}{x})= ke^{\frac{2y}{x}}$

b) $(x+2)^4 (1+ 2\frac{(y-2)}{x+2})= ke^{\frac{2(y-2)}{x+2}}$

c) $(x+2)^3 (1+ 2\frac{(y-2)}{x+2})= ke^{\frac{2(y-2)}{x+2}}$

d) None of these

Attempt:

I have expanded and checked but couldn't spot any exact differentials.

Secondly, it's not a homogeneous equation, so couldn't use $y = vx$.

How do I go about solving this problem?

Answer 1

hint: $x+y = (x+2) + (y-2)$ , and use $(A+B)^2 = A^2 + 2AB +B^2, A = x+2,B = y - 2$ to expand the numerator and simplify

Answer 2

If there is no typo in the equation, I do not think that a solution could be obtained.

The best I was able to do was, working with $x(y)$ instead of $y(x)$ $$x'=\frac{(y-2) (x+2)}{(x+y)^2}$$ Making $x=z-2$ gives $$z'=\frac{(y-2) z}{(z+y-2)^2} $$ which leads to an implicit equation $$\frac{z (3 \log (z)+\log (z+2 y-4)-1)-2 y+4}{4 z}=C$$ which looks impossible to solve.

Answer 3

Let $X=x+2$ and $Y=y-2$, so the given DE is equivalent to $$\dfrac{\mathrm d Y}{\mathrm d X}=\frac{(X+Y)^2}{X Y}$$ last DE is homogeneous, so can be transformed into a separable DE by making $Y =uX$ as follows $$\dfrac{\mathrm d Y}{\mathrm d X}=\frac{(X+Y)^2}{X Y}\qquad\iff\qquad u+X\frac{\mathrm d u}{\mathrm dX}=\frac{X^2(1+u)^2}{X^2u}$$ Then we get $$X\frac{\mathrm d u}{\mathrm d X}=\frac{1+2u}{u}\quad\implies\quad \frac{u}{2u+1}\dfrac{\mathrm d u}{\mathrm d X}=\frac1X$$ Integrating both sides respect to $X$ (supposing that $u$ depends on $X$) we get \begin{align*} \int\frac{u}{2u+1}\dfrac{\mathrm d u}{\mathrm d X}\mathrm d X&=\int\frac1X\mathrm dX\\[4pt] \int\left(\frac12-\frac{1/2}{2u+1}\right)\mathrm du&=\int\frac1X\mathrm dX\\[4pt] \frac12u-\frac14\ln|2u+1|&=\ln|X|+c_1 \end{align*} Last equality is equivalent to $$\frac{y-2}{x+2}-\frac12\ln\left|\frac{2(y-2)}{x+2}+1\right|=2\ln|x+2|+2c_1$$ Notice that this solution can be carried to the form b): \begin{align*} \frac{2(y-2)}{x+2}-4c_2&=4\ln|x+2|+\ln\left|\frac{2(y-2)}{x+2}+1\right|\\[4pt] k e^{\frac{2(y-2)}{x+2}}&=(x+4)^4\left[\frac{2(y-2)}{x+2}+1\right] \end{align*} where $k=\pm e^{-4c_1}$

Answer 4

$$\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}=\left(\dfrac{x+y}{x+2}\right)^2\dfrac{x+2}{y-2}$$ let $w=\dfrac{x+y}{x+2}$ then $y=w(x+2)-x$ and $$w'(x+2)+w-1=y'=w^2\dfrac{1}{w-1}$$ which is separable $$\dfrac{w-1}{2w-1}dw=\dfrac{dx}{x+2}$$ with integration the solution will be obtained.