Solve y"+4y=f(t) subject to y(0)=0,y′(0)=2 and f(t)={0t&lt;π1t≥π

Step 1 Given differential equation y"+4y=f(t) subject to y(0)=0,y′(0)=2 and f(t)={0t&lt;π1t≥πLf(t)=∫0∞f(t)e−stdt=∫0π(0)e−stdt+∫π∞(1)e−stdt=0+∫π∞(1)e−stdt=|e−st−s|π∞=|e−s(∞)−s−e−s(π)−s|=e−πtsStep 2Now taking Laplace transform on both sides of the given differential equationL{y″}+4L{y}=L{f(t)}s2Y(s)−s.y(0)−y′(0)+4Y(s)=e−πssApplying the conditionss2Y(s)−s.0−2+4Y(s)=e−πssY(s)(s2+4)=e−πss+2Y(s)=e−πss(s2+4)+2s2+4Now taking Inverse Laplace transform Step 3L−1{Y(s)}=L−1e−πss(s2+4)+2s2+4=L−1{e−πs1s(s2+4)}+2L−1{1s2+4}By partial fraction1s(s2+4)=14s−s4(s2+4)L−1{Y(s)}=L−1{e−πs(14s−s4(s2+4))}+2L−1{1s2+22}Step 4Apply the rule then L−1{e−asF(s)}=u(t−a)f(t−a) and from laplace transform tableL−1{14s−s4(s2+4)}=14−14cos⁡(2t)y(t)=u(1−π)(14−14cos⁡(2(t−π)))+2(12)sin⁡(2t)

"Solve y"+4y=f(t) subject to y(0)=0,y′(0)=2 and f(t)={0t<π1t≥π" solved and explained

abondantQ

Answered question

2021-02-12

Solve

y " + 4 y = f (t) subject to y (0) = 0, y^{'} (0) = 2 and

f (t) = {\begin{cases} 0 & t < π \\ 1 & t \geq π \end{cases}

Answer & Explanation

firmablogF

Skilled2021-02-13Added 92 answers

Step 1 Given differential equation $y " + 4 y = f (t)$ subject to $y (0) = 0, y^{'} (0) = 2$ and $f (t) = {\begin{cases} 0 & t < π \\ 1 & t \geq π \end{cases}$
$L f (t) = \int_{0}^{\infty} f (t) e^{- s t} d t$
$= \int_{0}^{π} (0) e^{- s t} d t + \int_{π}^{\infty} (1) e^{- s t} d t$
$= 0 + \int_{π}^{\infty} (1) e^{- s t} d t$
$= | \frac{e^{- s t}}{- s} |_{π}^{\infty}$
$= | \frac{e^{- s (\infty)}}{- s} - \frac{e^{- s (π)}}{- s} |$
$= \frac{e^{- π t}}{s}$
Step 2
Now taking Laplace transform on both sides of the given differential equation
$L {y^{″}} + 4 L {y} = L {f (t)}$
$s^{2} Y (s) - s . y (0) - y^{'} (0) + 4 Y (s) = \frac{e^{- π s}}{s}$
Applying the conditions
$s^{2} Y (s) - s .0 - 2 + 4 Y (s) = \frac{e^{- π s}}{s}$
$Y (s) (s^{2} + 4) = \frac{e^{- π s}}{s} + 2$
$Y (s) = \frac{e^{- π s}}{s (s^{2} + 4)} + \frac{2}{s^{2} + 4}$
Now taking Inverse Laplace transform Step 3
$L^{- 1} {Y (s)} = L^{- 1} \frac{e^{- π s}}{s (s^{2} + 4)} + \frac{2}{s^{2} + 4}$
$= L^{- 1} {e^{- π s} \frac{1}{s (s^{2} + 4)}} + 2 L^{- 1} {\frac{1}{s^{2} + 4}}$
By partial fraction
$\frac{1}{s (s^{2} + 4)} = \frac{1}{4 s} - \frac{s}{4 (s^{2} + 4)}$
$L^{- 1} {Y (s)} = L^{- 1} {e^{- π s} (\frac{1}{4 s} - \frac{s}{4 (s^{2} + 4)})} + 2 L^{- 1} {\frac{1}{s^{2} + 2^{2}}}$
Step 4
Apply the rule then $L^{- 1} {e^{- a s} F (s)} = u (t - a) f (t - a)$ and from laplace transform table
$L^{- 1} {\frac{1}{4 s} - \frac{s}{4 (s^{2} + 4)}} = \frac{1}{4} - \frac{1}{4} \cos (2 t)$
$y (t) = u (1 - π) (\frac{1}{4} - \frac{1}{4} \cos (2 (t - π))) + 2 (\frac{1}{2}) \sin (2 t)$

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