Question

The solution of the differential equation $\frac{dy}{dx}-\frac{y+3x}{\ln \left(y+3x\right)}+3=0$ is: (where $C$ is a constant of integration)

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

Answer 1

The correct option is B

$x-\frac{1}{2}\ln {\left(y+3x\right)}^2=C$

Explanation of the correct option.

Compute the required value.

We have the differential equation as,

$\frac{dy}{dx}-\frac{y+3x}{\ln \left(y+3x\right)}+3=0$

$\Rightarrow$ $\frac{dy}{dx}+3=\frac{y+3x}{\ln \left(y+3x\right)}$……………$\left(1\right)$

Let $\ln \left(y+3x\right)=t$

differentiate both side with respect to $x$.

$\Rightarrow \frac{1}{y+3x}\left(\frac{dy}{dx}+3\right)=\frac{dt}{dx}$

Substitute in equation $\left(1\right)$,

$\Rightarrow$ $\frac{dt}{dx}=\frac{1}{t}$

$\Rightarrow$ $tdt=dx$

Integrate both sides,

$\Rightarrow$ $\frac{t^2}{2}=x+C$

$\Rightarrow$ $\frac{1}{2}{\left(\ln \left(y+3x\right)\right)}^2=x+C$

Hence option $\left(B\right)$ is the correct option.