Question

The solution of differential equation $(3y-7x+7)dx+(7y-3x+3)dy=0$ is:
(where $C$ is constant of integration)

• A. $C={(x+y-2)}^2{(y-x+2)}^5$
• B. $C={(x+y-2)}^5{(y-x+2)}^2$
• C. $C={(x+y-1)}^2{(y-x+1)}^5$
• D. $C={(x+y-1)}^5{(y-x+1)}^2$

Answer 1

The correct option is D $C={(x+y-1)}^5{(y-x+1)}^2$

The given differential equation is $\frac{dy}{dx}=\frac{7x-3y-7}{-3x+7y+3}$
Substituting $x=X+h,y=Y+k,$ we obtain
$\frac{dY}{dX}=\frac{(7X-3Y)+(7h-3k-7)}{(-3X+7Y)+(-3h+7k+3)}\cdots \left(i\right)$
For homogenous we choose $h$ and $k$ such that,
$7h-3k-7=0$ and $-3h+7k+3=0$
This gives $h=1$ and $k=0.$ Under the above transformations, equation $\left(i\right)$ can be written as
$\frac{dY}{dX}=\frac{7X-3Y}{-3X+7Y}$
Let $Y=VX\Rightarrow \frac{dY}{dX}=V+X\frac{dV}{dX}$
So, the above differential equation tranforms to:
$V+X\frac{dV}{dX}=\frac{7-3V}{7V-3}\Rightarrow X\frac{dV}{dX}=\frac{7-7V^2}{7V-3}$
$\Rightarrow -7\frac{dX}{X}=\frac{7}{2}\cdot \frac{2V}{V^2-1}dV-\frac{3}{V^2-1}dV$
On integrating both sides, we get
$-7\ln |X|=\frac{7}{2}\ln |V^2-1|-\frac{3}{2}\ln \left|\frac{V-1}{V+1}\right|-\ln |C|\Rightarrow C={(V+1)}^5{(V-1)}^2{X}^7$

Now, on substituting the initial variables and simplifying, we have:
$\Rightarrow C={(x+y-1)}^5{(y-x+1)}^2$