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UPSC NDA I 2024 Mathematics Paper

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Question : 76 of 120

Marks: +1, -0

If

\frac{d y}{d x}=2 e^{x} y^{3}, y(0)=\frac{1}{2}

then what is 4y2(2 - ex) equal to ?

1
2
3
4

Solution:

Concept:

Solve the differential equation

\frac{dy}{dx}=2 e^{x} y^{3}

by separation of variables, use

y(0)=\frac{1}{2}

to find the constant, then compute

4y^{2}(2-e^{x})

.

Explanation:

Separate variables:

\frac{dy}{y^{3}} = 2 e^{x} dx

.

Rewrite as

y^{-3} dy = 2 e^{x} dx

.

Integrate both sides:

\int y^{-3} dy = \int 2 e^{x} dx

.

Result:

\frac{y^{-2}}{-2} = 2 e^{x} + C

.

Simplify:

-\frac{1}{2 y^{2}} = 2 e^{x} + C

.

Apply

y(0)=\frac{1}{2}

: substitute

x=0

,

y=\frac{1}{2}

.

-\frac{1}{2\left(\frac{1}{2}\right)^{2}} = 2 e^{0} + C

gives

-\frac{1}{2\cdot\frac{1}{4}} = 2 + C

→

-\frac{1}{\frac{1}{2}} = 2 + C

→

-2 = 2 + C

.

Thus

C = -4

.

Now

-\frac{1}{2 y^{2}} = 2 e^{x} - 4

.

Multiply by

-1

:

\frac{1}{2 y^{2}} = 4 - 2 e^{x} = 2(2 - e^{x})

.

Invert:

2 y^{2} = \frac{1}{2(2 - e^{x})}

. Multiply by

2

:

4 y^{2} = \frac{1}{2 - e^{x}}

.

Therefore

4 y^{2}(2 - e^{x}) = 1

.

Answer:

1

(Option A).

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