The solution of the differential equation dy/dx-(y+3x)/(log_e(y+3x))+3=0 is (where C is a constant of integration)

Correct answer: (A) x-1/2loge(y+3x)^2=C. dy/dx-(y+3x)/(ln(y+3x))+3=0 dy/dx+3=(y+3x)/(ln(y+3x)) d/dx(y+3x)=(y+3x)/(ln(y+3x)) ∫(ln(y+3x))/((y+3x))d(y+3x)=∫dx Let ln(y+3x)=t 1/((y+3x))d(y+3x)=dt ∫tdt=∫dx t^22=x+c ((ln(y+3x))^2)/2=x+c

Q34 / 75 · JEE Main 2020 (04 Sep Shift 2) · Mathematics

The solution of the differential equation $\frac{d y}{d x} - \frac{y + 3 x}{\log_{e} (y + 3 x)} + 3 = 0$ is
(where $C$ is a constant of integration)

Correct answer
A
$x - \frac{1}{2} {(\log_{e} (y + 3 x))}^{2} = C$
B
$x - \log_{e} (y + 3 x) = C$
C
$y + 3 x - \frac{1}{2} {(\log_{e} x)}^{2} = C$
D
$x - 2 \log_{e} (y + 3 x) = C$

Solution

\frac{d y}{d x} - \frac{y + 3 x}{\ln (y + 3 x)} + 3 = 0

\frac{d y}{d x} + 3 = \frac{y + 3 x}{\ln (y + 3 x)}

\frac{d}{d x} (y + 3 x) = \frac{y + 3 x}{\ln (y + 3 x)}

\int \frac{\ln (y + 3 x)}{(y + 3 x)} d (y + 3 x) = \int d x

Let

\ln (y + 3 x) = t

\frac{1}{(y + 3 x)} d (y + 3 x) = d t

\int tdt = \int dx

\frac{t^{2}}{2} = x + c

\frac{(\ln (y + 3 x))^{2}}{2} = x + c

Hence, the correct answer is option: 'A'.

The solution of the differential equation dydx-y+3xloge(y+3x)+3=0 is(where C is a constant of integration)

Solution

Relevant Questions

The solution of the differential equation $\frac{d y}{d x} - \frac{y + 3 x}{\log_{e} (y + 3 x)} + 3 = 0$ is
(where $C$ is a constant of integration)