Let y = y ( x ) be the solution of the differential equation d y d x + 2 y = f ( x ) , where f ( x ) = { 1 , x [ 0 , 1 ] 0 , otherwise . If y ( 0 ) = 0 , then y ( ln 2 ) is

Question

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 2 y = f (x)$ , where $f (x) =$ ${\begin{matrix} 1, & x \in [0, 1] \\ 0, & otherwise \end{matrix}$ . If $y (0) = 0$ , then $y (\ln 2)$ is

Smart Achievers · Accepted Answer

(4)

$When x \in [0, 1], then \frac{d y}{d x} + 2 y = 1 \Rightarrow y = \frac{1}{2} + C_{1} e^{- 2 x}$

$∵$ $y (0) = 0 \Rightarrow y (x) = \frac{1}{2} - \frac{1}{2} e^{- 2 x}$

$Here, y (1) = \frac{1}{2} - \frac{1}{2} e^{- 2 x} = \frac{e^{2} - 1}{2 e^{2}}$

$When x \notin [0, 1] then \frac{d y}{d x} + 2 y = 0 \Rightarrow C_{2} = \frac{e^{2} - 1}{2}$

$∵$ $y (1) = \frac{e^{2} - 1}{2} \Rightarrow \frac{e^{2} - 1}{2} = C_{2} e^{- 2} \Rightarrow C_{2} = \frac{e^{2} - 1}{2}$

$∵$ $y (x) = (\frac{e^{2} - 1}{2}) e^{- 2 x} \Rightarrow y (\frac{3}{2}) = \frac{e^{2} - 1}{2 e^{3}}$

Solution

Q.
Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 2 y = f (x)$ , where $f (x) =$ ${\begin{matrix} 1, & x \in [0, 1] \\ 0, & otherwise \end{matrix}$ . If $y (0) = 0$ , then $y (\ln 2)$ is

1 $\frac{e^{2} - 1}{4 e^{2}}$

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

3 $\frac{1}{4}$

4 $\frac{3}{8}$

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Solution

Q. Let y=y(x) be the solution of the differential equation dydx+2y=f(x), where f(x)={1,x∈[0,1]0,otherwise. If y(0)=0, then y(ln2) is

1 e2-14e2

2 e2-12e2

3 14

4 38