Question

JEE MAINS MATHS QUESTION SEPTEMBER 2020

Answer 1

Answer:

here is the explanation to the given question

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

$\displaystyle \: \frac{dy}{dx} - \frac{y - 3x}{ \ln(y - 3x)} = 3$

$\implies \: \: \displaystyle \: \frac{dy}{dx} - 3 = \frac{y - 3x}{ \ln(y - 3x)} .......(1)$

Let

$z = y - 3x$

Differentiating both sides with respect to x we get

$\: \: \displaystyle \: \frac{dz}{dx} =\frac{dy}{dx} - 3$

From Equation (1)

$\implies \: \: \displaystyle \: \frac{dz}{dx} = \frac{z}{ \ln \: z} .......(1)$

$\implies \: \: \displaystyle \: \: \frac{ \ln z}{z} \: {dz} = dx$

On integration

$\displaystyle \int\limits_{}^{} \frac{ \ln z}{z} \, dz = \int\limits_{}^{} \, dx \: \: \: ....(2)$

Let

$u \: = \ln z$

Differentiating both sides with respect z we get

$\displaystyle \: \frac{du}{dz} = \frac{1}{z}$

From Equation (2)

$\displaystyle \int\limits_{}^{} u \, du = \int\limits_{}^{} \, dx \: \: \:$

$\implies \: \displaystyle \: \frac{ {u}^{2} }{2} = x + c \: \: \: (where \: \: c = constant \: )$

Putting the value of u we get

$\displaystyle \: \frac{ { (\ln z)}^{2} }{2} = x + c \: \: \: (where \: \: c = constant \: )$

Putting the value of z we get

$\displaystyle \: \frac{ { \{\ln(y - 3x) \}}^{2} }{2} = x + c \: \: \: (where \: \: c = constant \: )$

So the required solution is

$\boxed{ \displaystyle \: \frac{ { \{\ln(y - 3x) \}}^{2} }{2} = x + c \: \: \: (where \: \: c = constant \: )}$