# Solve the differential equation (dy)/(dx)=(x+2y+3)/(2x+y+3) ?

Nov 29, 2017

See below.

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x + 2 y + 3}{2 x + y + 3} \Rightarrow \left(2 x + y + 3\right) \mathrm{dy} = \left(x + 2 y + 3\right) \mathrm{dx}$

Now we will do a series of variable transformations to get an amenable format in the differential equation

1) $\left\{\begin{matrix}x + 2 y = u \\ 2 x + y = v\end{matrix}\right. \Rightarrow \left\{\begin{matrix}\mathrm{dx} = \frac{1}{3} \left(- \mathrm{du} + 2 \mathrm{dv}\right) \\ \mathrm{dy} = \frac{1}{3} \left(2 \mathrm{du} - \mathrm{dv}\right)\end{matrix}\right.$

$\left(v + 3\right) \left(2 \mathrm{du} - \mathrm{dv}\right) = \left(u + 3\right) \left(- \mathrm{du} + 2 \mathrm{dv}\right)$

Now making the transformation

2) $\left\{\begin{matrix}U = u + 3 \\ V = v + 3\end{matrix}\right. \Rightarrow \left\{\begin{matrix}\mathrm{dU} = \mathrm{du} \\ \mathrm{dV} = \mathrm{dv}\end{matrix}\right.$

$V \left(2 \mathrm{dU} - \mathrm{dV}\right) = U \left(- \mathrm{dU} + 2 \mathrm{dV}\right)$ or

$2 V \mathrm{dU} - V \mathrm{dV} = - U \mathrm{dU} + 2 U \mathrm{dV}$

Now making

3) $\left\{\begin{matrix}\eta = {U}^{2} \\ \xi = {V}^{2}\end{matrix}\right. \Rightarrow \left\{\begin{matrix}\frac{1}{2} \frac{d \eta}{\sqrt{\eta}} = \mathrm{dU} \\ \frac{1}{2} \frac{d \xi}{\sqrt{\xi}} = \mathrm{dV}\end{matrix}\right.$

$2 \sqrt{\frac{\xi}{\eta}} d \eta - d \xi = - d \eta + 2 \sqrt{\frac{\eta}{\xi}} d \xi$ or

$\left(2 \sqrt{\frac{\xi}{\eta}} + 1\right) d \eta = \left(2 \sqrt{\frac{\eta}{\xi}} + 1\right) d \xi$

now introducing

4) $\eta = \lambda \xi \Rightarrow d \eta = \lambda d \xi + \xi d \lambda$

$\left(\frac{2}{\sqrt{\lambda}} + 1\right) \left(\lambda d \xi + \xi d \lambda\right) = \left(2 \sqrt{\lambda} + 1\right) d \xi$

or grouping variables

$\frac{d \xi}{\xi} + f \frac{\lambda}{\lambda f \left(\lambda\right) - 1} d \lambda = 0$

with $f \left(\lambda\right) = \frac{\frac{2}{\sqrt{\lambda}} + 1}{2 \sqrt{\lambda} + 1}$

After integration we obtain

$\log \xi + 3 \log \left(1 - \sqrt{\lambda}\right) - \log \left(1 + \sqrt{\lambda}\right) = {C}_{0}$ or

$\xi = \frac{{C}_{1} \left(1 + \sqrt{\lambda}\right)}{1 - \sqrt{\lambda}} ^ 3$ or

${V}^{2} = \frac{{C}_{1} \left(1 + \frac{U}{V}\right)}{1 - \frac{U}{V}} ^ 3$ and finally

${\left(2 x + y + 3\right)}^{2} = \frac{{C}_{1} \left(1 + \frac{x + 2 y + 3}{2 x + y + 3}\right)}{1 - \frac{x + 2 y + 3}{2 x + y + 3}} ^ 3$

An implicit form solution.

This can be reduced to

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

Nov 29, 2017

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

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x + 2 y + 3}{2 x + y + 3}$

$\left(2 x + y + 3\right) \cdot \mathrm{dy} = \left(x + 2 y + 3\right) \cdot \mathrm{dx}$

Now I solved $2 x + y + 3 = 0$ and $x + 2 y + 3 = 0$ equation system. From them, $x = y = - 1$

Hence I used $x = m - 1$, $y = p - 1$, $\mathrm{dx} = \mathrm{dm}$ and $\mathrm{dy} = \mathrm{dp}$ substitution, this differential equation became

$\left(2 m + p\right) \cdot \mathrm{dp} = \left(m + 2 p\right) \cdot \mathrm{dm}$

I used $m = p \cdot z$ and $\mathrm{dm} = p \mathrm{dz} + z \mathrm{dp}$ transformation, it became

$\left(2 p z + p\right) \cdot \mathrm{dp} = \left(p z + 2 p\right) \left(p \mathrm{dz} + z \mathrm{dp}\right) = 0$

$\left(2 z + 1\right) \cdot \mathrm{dp} = \left(z + 2\right) \left(p \mathrm{dz} + z \mathrm{dp}\right) = 0$

$\left(2 z + 1\right) \cdot \mathrm{dp} = p \left(z + 2\right) \cdot \mathrm{dz} + \left({z}^{2} + 2 z\right) \cdot \mathrm{dp}$

$\left({z}^{2} - 1\right) \cdot \mathrm{dp} + p \left(z + 2\right) \cdot \mathrm{dz} = 0$

$\frac{\mathrm{dp}}{p} + \frac{\left(z + 2\right) \cdot \mathrm{dz}}{{z}^{2} - 1} = 0$

$\frac{2 \mathrm{dp}}{p} + \frac{\left(2 z + 4\right) \cdot \mathrm{dz}}{\left(z + 1\right) \left(z - 1\right)} = 0$

$\frac{2 \mathrm{dp}}{p} + \frac{3 \mathrm{dz}}{z - 1} - \frac{\mathrm{dz}}{z + 1} = 0$

$2 L n p + 3 L n \left(z - 1\right) - L n \left(z + 1\right) = L n C$

$L n \left[{p}^{2} \cdot {\left(z - 1\right)}^{3} / \left(z + 1\right)\right] = L n C$

${p}^{2} \cdot {\left(z - 1\right)}^{3} / \left(z + 1\right) = C$

After using $m = p \cdot z$ and $z = \frac{m}{p}$ inverse transformation,

${p}^{2} \cdot {\left(\frac{m}{p} - 1\right)}^{3} / \left(\frac{m}{p} + 1\right) = C$

${\left(m - p\right)}^{3} / \left(m + p\right) = C$

After using $x = m - 1$, $y = p - 1$, $m = x + 1$ and $p = y + 1$ inverse transformation, I found

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