Question

Solve :
$\frac{dy}{dx}=\frac{6x-2y-7}{2x+3y-6}$

Answer 1

$\frac{dy}{dx}=\frac{6x-2y-7}{2x+3y-6}⟶\left(1\right)$

To solve this differential equation, we need to change the variables as $x=x+h$ and $y=y+k$ such that $6h-2k-7=0$ & $2h+3k-6=0$

Solving the two equations simultaneously we find the value of $(h,k)$ to be $(9/14,-11/7)$

$\therefore$ Substitute, $x=X+9/14$ and $y=Y-11/7$ and also $dy=dY$ and $dx=dX$

Equation $\left(1\right)$ transforms into

$\frac{dy}{dx}=\frac{6(x+9/14)-2(y-11/7)-7}{2(x+9/14)+3(y-11/7)-6}=\frac{6x-2y}{2x+3y}⟶\left(2\right)$

Now Substitute, $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}⟶\left(3\right)$

Putting $\left(3\right)$ in $\left(2\right)$ we get

$v+x\frac{dv}{dx}=\frac{6-2v}{2+3v}$

$\Rightarrow x\frac{dv}{dx}=\frac{6-2v-2v-{3v}^2}{2+3v}$

$\Rightarrow \frac{(2+3v)dv}{6-4v-{3v}^2}=\frac{dx}{x}$

$\Rightarrow \frac{-(2+3v)dv}{-6+4v+{3v}^2}=\frac{dx}{x}$

integrating both sides

$\Rightarrow \frac{-1}{2}\int \frac{(6v+4)dv}{{3v}^2+4v-6}dv=\int \frac{dx}{x}$

$\Rightarrow \frac{-1}{2}\int \frac{d({3v}^2+4v-6)}{{3v}^2+4v-6}dv=\int \frac{dx}{x}$

$\Rightarrow \frac{-1}{2}ln({3v}^2+4v-6)=lnx+lnc$

$\Rightarrow ln(3\frac{y^2}{x^2}+\frac{4y}{x}-\frac{6}{x})=-2ln\times c$

$\Rightarrow ln({3y}^2+4yx-{6x}^2)-ln{x}^2=-ln{x}^2-ln{c}^2$

$\Rightarrow {3y}^2+4yx-{6x}^2=A$ ; $A=c^{-2}$

$\Rightarrow 3{(y+11/7)}^2+4(x-9/14)(y+11/7)-6(x-9/14)=A$