Let the solution curve of the differential equation $x \frac{d y}{d x}-y=\sqrt{y^{2}+16 x^{2}}, y(1)=3$ be $y=y(x)$. Then $\mathrm{y}(2)$ is equal to :

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Differential Equations

Let the solution curve of the differential equation

x \frac{d y}{d x} - y = y^{2} + 16 x^{2}, y (1) = 3

y = y (x)

.
Then

y (2)

is equal to :

Answer: (A)

Substitute

y = v x

, so

\frac{d y}{d x} = v + x \frac{d v}{d x}

.

The equation becomes:

x (v + x \frac{d v}{d x}) - v x = v^{2} x^{2} + 16 x^{2}

Simplifying:

x^{2} \frac{d v}{d x} = x v^{2} + 16

This gives:

\frac{d v}{v ^{2} + 16} = \frac{d x}{x}

Integrating:

ln ∣ v + v^{2} + 16 ∣ = ln ∣ x ∣ + ln C

Substituting back

v = y / x

y + y^{2} + 16 x^{2} = C x^{2}

Using

y (1) = 3

3 + 9 + 16 = C \Rightarrow C = 8

For

x = 2

y + y^{2} + 64 = 32

Solving:

y = 15

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Single correct option

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