Question

$(3x-7x+7)dx+(7y-3x+3)dy=0$

Answer 1

i) The given one is a non homogeneous first order first degree differential equation. Solving of which is a lengthy process, involving two conversions, one of making homogeneous and another of making it as separable equation. As such it may not be possible for me to give entire solution. However, let me give brief, basing on which you may solve and obtain required end result.

ii) Rearranging the given one, $\frac{dy}{dx}=\frac{(7x-3y-7)}{(7y-3x+3)}$

iii) To make it homogeneous, let $x=X+h$ and $y=Y+k$

Differentiating with respect to respective variables,

$\frac{dx}{dX}=1$ and $\frac{dy}{dY}=1$

$\frac{\frac{dy}{dY}}{\frac{dx}{dX}}=1$;so,$\frac{dy}{dx}=\frac{dY}{dX}=1$

iv) So,$\frac{dy}{dx}=\frac{dY}{dX}=$ $\frac{(7X-3Y+7h-3k-7)}{(7Y-3X-3h+7k+3)}$

The above becomes homogeneous,when $7h-3k-7=0=-3h+7k+3$

Solving these,we have $h=1$ and $k=0$

Thus $x=X+1$and$y=Y$

As of these,$\frac{dY}{dX}=\frac{(7X-3Y)}{(7y-3X)}$

Now let $Y=VX$,where $V$ is another variable.

Differentiating,$\frac{dY}{dX}=X\ast \frac{dV}{dX}+V$

$X\ast \frac{dV}{dX}+V=\frac{(7X-3Y)}{(7y-3X)}$

Rearranging,simplifying and separating,

$\frac{(7V-3)\ast dV}{7\ast {(1-V)}^2}=\frac{dX}{X}$

$\frac{V\ast dV}{(1-V²)-\frac{3}{7}}\ast \frac{dy}{(1-V²)}=\frac{dX}{X}$

Integrating both sides,

$-\frac{1}{2}\ast ln|1-V²|-\frac{3}{14}\ast ln\left|\frac{(1+V)}{(1-V)}\right|=ln|X|+lnC$

$ln|X|+\frac{3}{14}\ast ln\left|\frac{(1+V)}{(1-V)}\right|+\frac{1}{2}\ast ln|1-V²|+lnC=0$

$CX\ast \sqrt{(1-V²)}\ast {\left[\frac{(1+V)}{(1-V)}\right]}^{\frac{3}{14}}=1$