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MHTCET 16 May 2024 Shift 1 Paper

Question : 143 of 150

Marks: +1, -0

\frac{d y}{d x}=y+3, y+3 > 0

y(0)=2

y(\log 2)

is equal to

Solution:

Concept:

The equation is a first-order differential equation solvable by separating the variables.

Explanation:

Given:

\frac{dy}{dx}=y+3

, where

y+3>0

and

y(0)=2

Separate the variables:

\frac{dy}{y+3}=dx

Integrate both sides:

\int \frac{dy}{y+3}=\int dx

\log(y+3)=x+C

Use the initial condition

y(0)=2

\log(2+3)=0+C

C=\log 5

Therefore:

\log(y+3)=x+\log 5

Taking antilogarithms:

y+3=5e^x

y=5e^x-3

Now find

y(\log 2)

y(\log 2)=5e^{\log 2}-3=5(2)-3=7

Answer:

7

, which is option C.

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