Question

Solve the given initial-value problem $y'' + 4y = g(x)$, $y(0) = 1$, $y'(0) = 1$, where $g(x) = \begin{cases} \sin(x), & 0 \le x \le \pi/2 \\ 0, & x > \pi/2 \end{cases}$ $g(x)$ is discontinuous. [Hint: Solve the problem on two intervals, and then find a solution so that $y$ and $y'$ are continuous at $x = \pi/2$.] $y(x) = \begin{cases} \text{ }, & 0 \le x \le \pi/2 \\ \text{ }, & x > \pi/2 \end{cases}$

Name: solve the given initial value problem y 4y gx y0 1 y0 1 where gx begincases sinx 0 le x le pi2 0 x pi2 endcases gx is discontinuous hint solve the problem on two intervals and then find a so 15789
Uploaded: 2022-01-03T04:13:17-08:00
Duration: 4 min 9 s
Channel: Nick Johnson
Description: solve the given initial value problem y 4y gx y0 1 y0 1 where gx begincases sinx 0 le x le pi2 0 x pi2 endcases gx is discontinuous hint solve the problem on two intervals and then find a so 15789

          Solve the given initial-value problem
$y'' + 4y = g(x)$, $y(0) = 1$, $y'(0) = 1$, where
$g(x) = \begin{cases} \sin(x), & 0 \le x \le \pi/2 \\ 0, & x > \pi/2 \end{cases}$
$g(x)$ is discontinuous. [Hint: Solve the problem on two intervals, and then find a solution so that $y$ and $y'$ are continuous at $x = \pi/2$.]
$y(x) = \begin{cases} \text{ }, & 0 \le x \le \pi/2 \\ \text{ }, & x > \pi/2 \end{cases}$

solve the given initial value problem y 4y gx y0 1 y0 1 where gx begincases sinx 0 le x le pi2 0 x pi2 endcases gx is discontinuous hint solve the problem on two intervals and then find a so 15789

Added by Timothy O.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Nick Johnson

01/03/2022

Step 1

First, let's find the general solution to the homogeneous equation $y'' + 4y = 0$. The characteristic equation is $r^2 + 4 = 0$, which gives $r^2 = -4$, so $r = \pm 2i$. The general solution to the homogeneous equation is $y_h(x) = c_1 \cos(2x) + c_2 Show more…

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Solve the given initial-value problem $y'' + 4y = g(x)$, $y(0) = 1$, $y'(0) = 1$, where $g(x) = \begin{cases} \sin(x), & 0 \le x \le \pi/2 \\ 0, & x > \pi/2 \end{cases}$ $g(x)$ is discontinuous. [Hint: Solve the problem on two intervals, and then find a solution so that $y$ and $y'$ are continuous at $x = \pi/2$.] $y(x) = \begin{cases} \text{ }, & 0 \le x \le \pi/2 \\ \text{ }, & x > \pi/2 \end{cases}$