Question

The solution of the differential equation $\frac{dy}{dx}-\frac{y+3x}{{\log }_e(y+3x)}+3=0$ is
(where $c$ is a constant of integration)

• A. $x-2{\log }_e(y+3x)=c$
• B. $x-{\log }_e(y+3x)=c$
• C. $x-\frac{1}{2}{\left({\log }_e\right(y+3x\left)\right)}^2=c$
• D. $y+3x-\frac{1}{2}{\left({\log }_{e}x\right)}^2=c$

Answer 1

The correct option is C $x-\frac{1}{2}{\left({\log }_e\right(y+3x\left)\right)}^2=c$

Given : $\frac{dy}{dx}-\frac{y+3x}{{\log }_e(y+3x)}+3=0$
$\Rightarrow \frac{dy}{dx}=\frac{y+3x}{{\log }_e(y+3x)}-3\cdots \left(1\right)$

Let $\ln (y+3x)=z$
Differentiating with respect to $x$,
$\frac{1}{y+3x}\cdot (\frac{dy}{dx}+3)=\frac{dz}{dx}$
From equation $\left(1\right)$, we get
$\frac{1}{z}=\frac{dz}{dx}\Rightarrow \int zdz=\int dx+C\Rightarrow \frac{z^2}{2}=x+C\Rightarrow \frac{1}{2}\left(\ln \right(y+3x){)}^2=x+C\therefore x-\frac{1}{2}(\ln (y+3x){)}^2=c$
where $c=-C$