Question

Find the solution of ${(3x+4y-5)}^{{}^2}\frac{dy}{dx}=a^2$

• A. $4y+\lambda =\frac{2a}{\sqrt{3}}{\tan }^{-1}\frac{(3x+4y-5)\sqrt{3}}{2a}$
• B. $2y+\lambda =\frac{5a}{\sqrt{3}}{\tan }^{-1}\frac{(3x+4y-5)\sqrt{3}}{2a}$
• C. $2y+\lambda =\frac{2a}{\sqrt{3}}{\tan }^{-1}\frac{(3x+4y-5)\sqrt{3}}{2a}$
• D. $4y+\lambda =\frac{5a}{\sqrt{3}}{\tan }^{-1}\frac{(3x+4y-5)\sqrt{3}}{2a}$

Answer 1

The correct option is A $4y+\lambda =\frac{2a}{\sqrt{3}}{\tan }^{-1}\frac{(3x+4y-5)\sqrt{3}}{2a}$

${(3x+4y-5)}^2\frac{dy}{dx}=a^2$
Let $3x+4y-5=v$
$3+4\frac{dy}{dx}=\frac{dv}{dx}$
$\Rightarrow \frac{dy}{dx}=\frac{1}{4}(\frac{dv}{dx}-3)$
Substituting these values in given differential equation
$v^2[\frac{dv}{dx}-3]=4a^2$
$\Rightarrow \frac{dv}{dx}=\frac{4a^2+3v^2}{v^2}$
$\Rightarrow \frac{v^2}{4a^2+3v^2}dv=dx$
$\Rightarrow \frac{v^2}{v^2+\frac{4}{3}{a}^2}dv=3dx$
Integrating both sides,
$\int (1-\frac{4}{3}\frac{a^2}{v^2+\frac{4}{3}{a}^2})dv=3x+C$
$\Rightarrow v-\frac{2a}{\sqrt{3}}{\tan }^{-1}\left(\frac{\sqrt{3}v}{2a}\right)=3x+C$
$\Rightarrow 4y-5=\frac{2a}{\sqrt{3}}{\tan }^{-1}\left(\frac{\sqrt{3}(3x+4y-5)}{2a}\right)+C$
$\Rightarrow 4y+\lambda =\frac{2a}{\sqrt{3}}{\tan }^{-1}\left(\frac{\sqrt{3}(3x+4y-5)}{2a}\right)$ where $\lambda =-5-C$