JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Differential Equations

Let the curve

y = y (x)

be the solution of the differential equation,

\frac{d y}{d x} = 2 (x + 1)

. If the numerical value of area bounded by the curve

y = y (x)

and

x

-axis is

\frac{4 8}{3}

, then the value of

y (1)

is equal to ____

Answer: 2

Solution

Integrating,

y = (x + 1)^{2} + c

. A finite region between the parabola and the

x

-axis exists only when

c < 0

, so write

c = - k^{2}

with

k > 0

; the curve meets the axis at

x = - 1 \pm k

, and with

u = x + 1

the enclosed area is

\int_{- k}^{k} (k^{2} - u^{2}) d u = \frac{4 k ^{3}}{3}

Setting

\frac{4 k ^{3}}{3} = \frac{4 8}{3}

gives

k^{3} = 22

, so

k = 2

and

c = - 2

. Hence

y (1) = (1 + 1)^{2} - 2 = 2

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Integer answer

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