SATHEE: Differential Equations 2 Question 20

Differential Equations 2 Question 20

21. If $y (t)$ is a solution of $(1 + t) \frac{d y}{d t} - t y = 1$ and $y (0) = - 1$ , then $y (1)$ is equal to

$(2003, 1$ M)

(a) $- 1 / 2$

(b) $e + 1 / 2$

(d) $1 / 2$

Objective Questions II

(One or more than one correct option)

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Answer:

Correct Answer: 21. (a)

Solution:

Given, $\frac{d y}{d t} - \frac{t}{1 + t} y = \frac{1}{(1 + t)}$ and $y (0) = - 1$

Which represents linear differential equation of first order.

$∴ I F = e^{\int - \frac{t}{1 + t} d t} = e^{- t + \log (1 + t)} = e^{- t} \cdot (1 + t)$

Required solution is,

$\begin{array}{lc} y e^{- t} (1 + t) = \int \frac{1}{1 + t} \cdot e^{- t} (1 + t) d t + c = \int e^{- t} d t + c \\ \Rightarrow & y e^{- t} (1 + t) = - e^{- t} + c \end{array}$

$\begin{aligned} Since, \\ y (0) = - 1 \\ \Rightarrow - 1 \cdot e^{0} (1 + 0) = - e^{0} + c \\ c = 0 \\ ∴ y = - \frac{1}{(1 + t)} \Rightarrow y (1) = - \frac{1}{2} \end{aligned}$

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Differential Equations 2 Question 20

21. If y(t) is a solution of (1+t)dydt−ty=1 and y(0)=−1, then y(1) is equal to

Objective Questions II

Answer:

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21. If $y (t)$ is a solution of $(1 + t) \frac{d y}{d t} - t y = 1$ and $y (0) = - 1$ , then $y (1)$ is equal to