If $(t + 1) - ty = 1$ and $y (0) = - 1$ , then at t=1, the solution of the differential equation is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$\log_{e}^{2} + \frac{1}{2}$

$\frac{- 1}{2}$

$\frac{5}{4}$

$e \frac{- 1}{2}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} G i v e n \frac{d y}{d x} - \frac{t}{1 + t} y = \frac{1}{t + 1} \\ I . F = e^{\int \frac{- t}{t + 1} d t} = e^{\int (\frac{1}{t + 1} - 1) d t} \\ = e^{\log (t + 1) - t} \\ = (t + 1) e^{- t} \\ S o k u t i o n i s \\ y (I . F) = \int \frac{1}{t + 1} (I . F) d t \\ \Rightarrow y (t + 1) e^{- t} = \int \frac{1}{t + 1} (t + 1) e^{- t} d t \\ = \int e^{- t} d t \end{array}$

$\begin{array}{l} = - e^{- t} + c \\ ∴ y (t + 1) = - 1 + c e^{t} \\ p u t t = 0 \\ \Rightarrow y (0) .1 = - 1 + c \\ c = 0 \\ ∴ y (t + 1) = - 1 \\ p u t t = 1 \\ y = \frac{- 1}{2} \end{array}$