If $\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{1+\mathrm{x}-\mathrm{y}^{2}}{\mathrm{y}}, \mathrm{x}(1)=1$, then $5 \mathrm{x}(2)$ is equal to :

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Differential Equations

\frac{dx}{dy} = \frac{1 + x - y ^{2}}{y}, x (1) = 1

, then

5 x (2)

is equal to :

Answer: 5

Rewrite

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

\frac{d x}{d y} - \frac{x}{y} = \frac{1 - y ^{2}}{y} = \frac{1}{y} - y,

a linear equation in

x

as a function of

y

, with integrating factor

I F = e^{\int - \frac{1}{y} d y} = e^{- l n y} = \frac{1}{y} .

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

Integrating:

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

Using

x (1) = 1

1 = - 1 - 1 + C ⟹ C = 3

x (y) = - 1 - y^{2} + 3 y .

x (2) = - 1 - 4 + 6 = 1.

5 x (2) = 5.

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Integer answer

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