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) $e+1 / 2$
(c) $e-1 / 2$
(d) $1 / 2$
Objective Questions II
(One or more than one correct option)
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Answer:
Correct Answer: 21. (a)
Solution:
- Given, $\frac{d y}{d t}-\frac{t}{1+t} \quad y=\frac{1}{(1+t)}$ and $y(0)=-1$
Which represents linear differential equation of first order.
$$ \therefore \quad IF=e^{\int-\frac{t}{1+t} d t}=e^{-t+\log (1+t)}=e^{-t} \cdot(1+t) $$
Required solution is,
$$ \begin{array}{lc} y e^{-t}(1+t)=\int \frac{1}{1+t} \cdot e^{-t}(1+t) d t+c=\int e^{-t} d t+c \\ \Rightarrow & y e^{-t}(1+t)=-e^{-t}+c \end{array} $$
$$ \begin{aligned} & \text { Since, } \\ & y(0)=-1 \\ & \Rightarrow \quad-1 \cdot e^{0}(1+0)=-e^{0}+c \\ & c=0 \\ & \therefore \quad y=-\frac{1}{(1+t)} \Rightarrow y(1)=-\frac{1}{2} \end{aligned} $$
