Question

# Let $$y = y(x)$$ be the solution of the differential equation $$\dfrac {dy}{dx} + 2y = f(x)$$, where $$f(x) = \left\{\begin{matrix} 1, & x\in [0, 1]\\ 0, & otherwise\end{matrix}\right.$$If $$y(0) = 0$$, then $$y\left (\dfrac {3}{2}\right )$$ is

A
e212e3
B
e21e3
C
12e
D
e2+12e4

Solution

## The correct option is D $$\dfrac {e^{2} - 1}{2e^{3}}$$Solving the initial value problem, we get $$y = \dfrac{1}{2} - \dfrac{1}{2}e^{-2x}$$ when $$x \in [0, 1]$$. We can check this by substituting this in the differential equation and checking the initial value.So, $$y(1) = \dfrac{1-e^{-2}}{2} = \dfrac{e^2-1}{2e^2} ... (1)$$Now, for $$x \in (1, \infty)$$, we have $$e^{2x}y = c_2$$ (solving the differential equation separately for this interval)Using the condition found above in $$(1)$$, we have $$c_2 = \dfrac{e^2-1}{2}$$. That gives $$y = \dfrac{e^2-1}{2}e^{-2x}$$ for $$x \in (1, \infty)$$So, for $$x=\dfrac{3}{2}$$, we get $$y = \dfrac{e^2-1}{2e^3}$$. So, the correct answer is option A.Mathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More