Question

Which of the following is the solution of the DE $\frac{\mathrm{dy}}{\mathrm{dx}}=(3x+2y+4{)}^2$?

• A. $\frac{1}{\sqrt{6}}{\tan }^{-1}\left(\sqrt{\frac{2}{3}}(3x+2y+4)\right)=x+c$
• B. $\frac{1}{\sqrt{6}}{\tan }^{-1}\left(\sqrt{\frac{2}{3}}(3x+2y)\right)=x+c$
• C. $\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\sqrt{\frac{2}{3}}(3x+2y+4)\right)=x+c$
• D. $\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\sqrt{\frac{2}{3}}(3x+2y)\right)=x+c$

Answer 1

The correct option is A $\frac{1}{\sqrt{6}}{\tan }^{-1}\left(\sqrt{\frac{2}{3}}(3x+2y+4)\right)=x+c$

$\mathrm{This}\mathrm{is}\mathrm{reducible}\mathrm{to}\mathrm{variable}\mathrm{separable}\mathrm{form}.\text{Putting 3x+2y+4=v}\text{3+2}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dv}}{\mathrm{dx}}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\left(\frac{\mathrm{dv}}{\mathrm{dx}}-3\right)\mathrm{So},\mathrm{the}\mathrm{equation}\mathrm{becomes}\frac{1}{2}\left(\frac{\mathrm{dv}}{\mathrm{dx}}-3\right)=v^2\frac{\mathrm{dv}}{\mathrm{dx}}=2v^2+3\int \frac{\mathrm{dv}}{2v^2+3}=\int \mathrm{dx}+c\int \frac{\mathrm{dv}}{{\left(\sqrt{3}\right)}^2+{\left(\sqrt{2}v\right)}^2}=x+c\frac{1}{\sqrt{3}}\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\sqrt{2}v}{\sqrt{3}}\right)=x+c\mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}v\frac{1}{\sqrt{6}}{\tan }^{-1}\left(\sqrt{\frac{2}{3}}(3x+2y+4)\right)=x+c$