Suppose $y=y(x)$ be the solution curve to the differential equation $\frac{d y}{d x}-y=2-e^{-x}$ such that $\lim _{x \rightarrow \infty} y(x)$ is finite. If $a$ and $b$ are respectively the $x$ and y-intercepts of the tangent to the curve at $\mathrm{x}=0$, then the value of $\mathrm{a}-4 \mathrm{~b}$ is equal to ______.

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Differential Equations

Suppose

y = y (x)

be the solution curve to the differential equation

\frac{d y}{d x} - y = 2 - e^{- x}

such that

lim_{x \to \infty} y (x)

is finite. If

a

and

b

are respectively the

x

and y-intercepts of the tangent to the curve at

x = 0

, then the value of

a - 4 b

is equal to ______.

Answer: 3

The differential equation

\frac{d y}{d x} - y = 2 - e^{- x}

has integrating factor

e^{- x}

. Multiplying and integrating:

y e^{- x} = \int (2 e^{- x} - e^{- 2 x}) d x = - 2 e^{- x} + \frac{e ^{- 2 x}}{2} + C

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

For

lim_{x \to \infty} y (x)

to be finite, we need

C = 0

. Thus

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

.

At

x = 0

y (0) = - 2 + \frac{1}{2} = - \frac{3}{2}

and

\frac{d y}{d x}_{x = 0} = - \frac{1}{2}

.

The tangent at

(0, - \frac{3}{2})

with slope

- \frac{1}{2}

is:

y + \frac{3}{2} = - \frac{1}{2} (x - 0) \Rightarrow x + 2 y = - 3

The intercepts are

a = - 3

(x-intercept) and

b = - \frac{3}{2}

(y-intercept).

Therefore:

a - 4 b = - 3 - 4 (- \frac{3}{2}) = - 3 + 6 = 3

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Integer answer