use the Laplace transform to solve the given initial-value problem.y"+y=f(t)y(0)=0,y′(0)=1 wheref(t)={10≤t&lt;π20t≥π2

Step 1Given, the initial value problemy"+y=f(t)y(0)=0,y′(0)=1 wheref(t)={10≤t&lt;π20t≥π2step 2The Laplace transform of the two derivativesL{y′}=sY(s)−y(0)L{y}=s2Y(s)−sy(0)−y′(0)L{1}=1sL{eat}=1s−aWhere y(0) and y’(0) are initial conditions.Step 3Now,when 0≤t≤π2y"+y′=1,y(0)=0.y′(0)=1s2Y(s)−sy(0)−y′(0)+sY(s)−y(0)=1{s2+s}Y(s)−s⋅0−1−0=1{s2+s}Y(s)=2Y(s)=2s2+s=2s(s+1)=2{1s−1s+1}Y(s)=2{1−e−t}when t≥π2y"+y′=0,y(0)=0,y′(0)=1s2Y(s)−sy(0)−y′(0)+sY(s)−y(0)=1{s2+s}Y(s)−s⋅0−1−0=0{s2+s}Y(s)=1Y(s)=1s2+s=1s(s+1)={1s−1s+1}Y(s)={1−e−t}Step 4Hence the solution is

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bobbie71G

Answered question

2021-03-04

use the Laplace transform to solve the given initial-value problem.
$y " + y = f (t)$
$y (0) = 0, y^{'} (0) = 1$ where
$f (t) = {\begin{matrix} 1 & 0 \leq t < \frac{π}{2} \\ 0 & t \geq \frac{π}{2} \end{matrix}$

Answer & Explanation

SchulzD

Skilled2021-03-05Added 83 answers

Step 1
Given, the initial value problem
$y " + y = f (t)$
$y (0) = 0, y^{'} (0) = 1$ where
$f (t) = {\begin{matrix} 1 & 0 \leq t < \frac{π}{2} \\ 0 & t \geq \frac{π}{2} \end{matrix}$
step 2
The Laplace transform of the two derivatives
$L {y^{'}} = s Y (s) - y (0)$
$L {y} = s^{2} Y (s) - s y (0) - y^{'} (0)$
$L {1} = \frac{1}{s}$
$L {e^{a t}} = \frac{1}{s - a}$
Where y(0) and y’(0) are initial conditions.
Step 3
Now,
when $0 \leq t \leq \frac{π}{2}$
$y " + y^{'} = 1, y (0) = 0 . y^{'} (0) = 1$
$s^{2} Y (s) - s y (0) - y^{'} (0) + s Y (s) - y (0) = 1$
${s^{2} + s} Y (s) - s \cdot 0 - 1 - 0 = 1$
${s^{2} + s} Y (s) = 2$
$Y (s) = \frac{2}{s^{2} + s} = \frac{2}{s (s + 1)} = 2 {\frac{1}{s} - \frac{1}{s + 1}}$
$Y (s) = 2 {1 - e^{- t}}$
when $t \geq \frac{π}{2}$
$y " + y^{'} = 0, y (0) = 0, y^{'} (0) = 1$
$s^{2} Y (s) - s y (0) - y^{'} (0) + s Y (s) - y (0) = 1$
${s^{2} + s} Y (s) - s \cdot 0 - 1 - 0 = 0$
${s^{2} + s} Y (s) = 1$
$Y (s) = \frac{1}{s^{2} + s} = \frac{1}{s (s + 1)} = {\frac{1}{s} - \frac{1}{s + 1}}$
$Y (s) = {1 - e^{- t}}$
Step 4
Hence the solution is

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