JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

The solution of

2\left( {y + 3} \right) - xy\frac{{dy}}{{dx}} = 0

with

y = - 2

, where

x = 1

, is

View answer and explanation

Correct option: D

{x^2}{\left( {y + 3} \right)^3} = {e^{y + 2}}

The correct answer is D (

{x^2}{\left( {y + 3} \right)^3} = {e^{y + 2}}

) because it is the solution to the given differential equation

2\left( {y + 3} \right) - xy\frac{{dy}}{{dx}} = 0

with the initial condition

y = -2

when

x = 1

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