JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Differential Equations

Which of the following is true for

y (x)

that satisfies the differential equation

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

;

y (0) = 0

Answer: (D)

Solution

The right-hand side factors:

x y - 1 + x - y = x (y + 1) - (y + 1) = (x - 1) (y + 1)

, so the equation separates as

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

Integrating,

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

, and

y (0) = 0

gives

C = 0

. Hence

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

x = 1

the exponent is

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

, so

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

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