Question

The solution of the differential equation $\frac{dy}{dx}=\frac{3y-7x-3}{3x-7y+7}$ is

• A. $(y-x-2{)}^5(y+x-5{)}^7=c$
• B. $(y-x-5{)}^2(y+x-1{)}^7=c$
• C. $(y-x-7{)}^2(y+x-5)=c$
• D. $(y-x-1{)}^2(y+x-1{)}^5=c$

Answer 1

Answer: (D)

$(y-x-1{)}^2(y+x-1{)}^5=c$

$\frac{dy}{dx}=\frac{3y-7x-3}{3x-7y+7}$ put $y=Y+1$ and $x=X$

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

$\frac{dY}{dX}=\frac{3Y-7X}{3X-7Y}$ put $Y=vX$ $\Rightarrow \frac{dY}{dX}=v+X\frac{dv}{dX}$

$\Rightarrow V+X\frac{dv}{dX}=\frac{3v-7}{3-7v}\Rightarrow \frac{3v-7v^2-3v+7}{(3-7v)}=-X\frac{dv}{dX}$

$X\frac{dv}{dX}=\frac{7(v+1)(v-1)}{(3-7v)}$ $\Rightarrow \frac{dv(3-7v)}{(v+1)(v-1)}=7\frac{dx}{X}$

$\frac{3-7v}{(v+1)(v-1)}=\frac{7-4-7v}{(v+1)(v-1)}=\frac{7(1-v)}{(v+1)(v-1)}-\frac{4}{(v+1)(v-1)}$

$=\frac{-7}{(v+1)}-\frac{4}{(v+1)(v-1)}$

$\Rightarrow \frac{3-7v}{(v+1)(v-1)}=\frac{-7}{(v+1)}-\frac{4}{(v+1)(v-1)}=\frac{-7}{(v+1}-\frac{4}{2}[\frac{1}{(v-1)}-\frac{1}{(v+1)}]=\frac{-7+2}{(v+1)}[\frac{1}{(v+1)}-\frac{1}{(v-1)}]$

$\frac{3-7v}{(v+1)(v-1)}=[\frac{-5}{(v+1)}-\frac{2}{(v-1)}]$

$\Rightarrow dv[\frac{-5}{(v+1)}-\frac{2}{(v-1)}]$ $7\frac{dx}{x}$

$\Rightarrow |(v+1{)}^5(v-1{)}^2|=7|n||x|$

$\Rightarrow X^7(v+1{)}^5(v-1{)}^2=c$

$\Rightarrow x^7\frac{(Y+x{)}^5(Y-x{)}^2}{x^7}=c$

$\Rightarrow (y+x{)}^5(y-x{)}^2=c$

$\Rightarrow (y-1+x{)}^5(y-1-x{)}^2=c$

$\Rightarrow (y+x-1{)}^5(y-x-1{)}^2=c$