Answer to Question #150169 in Differential Equations for kudachi

Question

(2x+3y)dx-4(x+1)dy=0

Answer 1

Let us solve the differential equation $(2x+3y)dx-4(x+1)dy=0$. It follows that $x=-1$ is a solution. Further, we will assume that $x\ne -1$. So, in this case the equation is equivalent to the equation

$\frac{dy}{dx}=\frac{2x+3y}{4(x+1)}$.

Let us solve the system $\begin{cases}2x+3y=0\\x+1=0\end{cases}$ which is equivalent to $\begin{cases}x=-1\\y=\frac{2}{3}\end{cases}$.

Let us introduce a substitution $\begin{cases}x=t-1\\y=z+\frac{2}{3}\end{cases}$.

Then we have he differential equation $\frac{dz}{dt}=\frac{2(t-1)+3(z+\frac{2}{3})}{4(t-1+1)}=\frac{2t+3z}{4t}$.

Let us introduce a substitution $z=t\cdot u(t)$. Then $dz=u\cdot dt+t\cdot du$ and thus

$\frac{u\cdot dt+t\cdot du}{dt}=\frac{2t+3tu}{4t}= \frac{2+3u}{4}$

$u+t\frac{du}{dt}= \frac{2+3u}{4}$

$t\frac{du}{dt}= \frac{2+3u}{4}-u=\frac{2-u}{4}$

If $u=2$, then $z=2t$, $y-\frac{2}{3}=2x+2$, and therefore, $y=2x+2+\frac{2}{3}=2x+\frac{8}{3}$. Since $(2x+3y)dx-4(x+1)dy=(2x+3(2x+\frac{8}{3}))dx-4(x+1)2dx=(8x+8)dx-8(x+1)dx=0,$ we conclude that $y=2x+\frac{8}{3}$ is a solution of the differential equation.

Let $u\ne 2$. Then

$\frac{4du}{2-u} =\frac{dt}{t}$

$\int\frac{4du}{2-u} =\int\frac{dt}{t}$

$-4\ln|2-u|+\ln|C|=\ln|t|$

$\ln|C|=\ln|t|+ \ln(2-u)^4=\ln|t(2-u)^4|$

$t(2-u)^4=C$

$t(2-\frac{z}{t})^4=C$

$(x+1)(2-\frac{y-\frac{2}{3}}{x+1})^4=C$

$(x+1)(\frac{2x-y+\frac{8}{3}}{x+1})^4=C$

$(2x-y+\frac{8}{3})^4=C(x+1)^3$

If $C=0$, then we have the solution $y=2x+\frac{8}{3}$.

Therefore, we have the following solution of the differentail equation $(2x+3y)dx-4(x+1)dy=0$:

$(2x-y+\frac{8}{3})^4=C(x+1)^3$ and $x=-1.$