Question

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + 2y = f(x)$, where $f(x) = \begin{cases} 1, \quad x \in [0, 1] \\ 0, \quad \text{otherwise} \end{cases}$

If $y(0) = 0$, then $y(\frac{3}{2})$ is:

• A. $\frac{e^2 - 1}{2e^3}$
• B. $\frac{e^2 - 1}{e^3}$
• C. $\frac{e^2 + 1}{2e^4}$
• D. $\frac{1}{2e}$

Answer 1

Answer: (A)

(A) $\frac{e^2 - 1}{2e^3}$

Answer 2

The given differential equation is a first-order linear differential equation: $\frac{dy}{dx} + 2y = f(x)$ where $f(x)$ is a piecewise function: $f(x) = \begin{cases} 1, & x \in [0, 1] \\ 0, & \text{otherwise} \end{cases}$ The initial condition is $y(0) = 0$. We need to find $y\left(\frac{3}{2}\right)$.

The differential equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)$, with $P(x) = 2$. The integrating factor (IF) is $e^{\int P(x) dx} = e^{\int 2 dx} = e^{2x}$.

We solve the differential equation in two intervals:

Case 1: $x \in [0, 1]$ In this interval, $f(x) = 1$. The differential equation becomes: $\frac{dy}{dx} + 2y = 1$ Multiplying by the integrating factor $e^{2x}$: $e^{2x} \frac{dy}{dx} + 2e^{2x} y = e^{2x}$ The left side is the derivative of $(y e^{2x})$: $\frac{d}{dx}(y e^{2x}) = e^{2x}$ Integrating both sides with respect to $x$: $y e^{2x} = \int e^{2x} dx + C_1$ $y e^{2x} = \frac{e^{2x}}{2} + C_1$ Dividing by $e^{2x}$ to find $y(x)$: $y(x) = \frac{1}{2} + C_1 e^{-2x}$ Now, apply the initial condition $y(0) = 0$: $0 = \frac{1}{2} + C_1 e^{-2(0)}$ $0 = \frac{1}{2} + C_1$ $C_1 = -\frac{1}{2}$ So, for $x \in [0, 1]$, the solution is: $y(x) = \frac{1}{2} - \frac{1}{2} e^{-2x}$

Case 2: $x > 1$ In this interval, $f(x) = 0$. The differential equation becomes: $\frac{dy}{dx} + 2y = 0$ Multiplying by the integrating factor $e^{2x}$: $e^{2x} \frac{dy}{dx} + 2e^{2x} y = 0$ $\frac{d}{dx}(y e^{2x}) = 0$ Integrating both sides with respect to $x$: $y e^{2x} = C_2$ $y(x) = C_2 e^{-2x}$ To ensure that $y(x)$ is a continuous solution, the value of $y(x)$ at $x=1$ must be the same for both expressions. From Case 1, $y(1) = \frac{1}{2} - \frac{1}{2} e^{-2(1)} = \frac{1}{2} - \frac{1}{2} e^{-2}$. From Case 2, $y(1) = C_2 e^{-2(1)} = C_2 e^{-2}$. Equating these two values: $C_2 e^{-2} = \frac{1}{2} - \frac{1}{2} e^{-2}$ $C_2 = \left(\frac{1}{2} - \frac{1}{2} e^{-2}\right) e^2$ $C_2 = \frac{1}{2} e^2 - \frac{1}{2} e^{-2} e^2$ $C_2 = \frac{e^2 - 1}{2}$ So, for $x > 1$, the solution is: $y(x) = \left(\frac{e^2 - 1}{2}\right) e^{-2x}$

Finally, we need to find $y\left(\frac{3}{2}\right)$. Since $\frac{3}{2} = 1.5$, which is greater than 1, we use the solution from Case 2: $y\left(\frac{3}{2}\right) = \left(\frac{e^2 - 1}{2}\right) e^{-2 \cdot \frac{3}{2}}$ $y\left(\frac{3}{2}\right) = \left(\frac{e^2 - 1}{2}\right) e^{-3}$ $y\left(\frac{3}{2}\right) = \frac{e^2 - 1}{2e^3}$

Comparing this result with the given options: (A) $\frac{e^2 - 1}{2e^3}$ (B) $\frac{e^2 - 1}{e^3}$ (C) $\frac{e^2 + 1}{2e^4}$ (D) $\frac{1}{2e}$

The calculated value matches option (A).