 # Solve the initial value problem x^{(4)}-5x"+4x=1-u_x(t), x(0)=x'(0)=x"(0)=x'''(0)=0 Tazmin Horton 2021-01-10 Answered
Solve the initial value problem
${x}^{\left(4\right)}-5x"+4x=1-{u}_{x}\left(t\right),$
$x\left(0\right)={x}^{\prime }\left(0\right)=x"\left(0\right)={x}^{‴}\left(0\right)=0$
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Step 1
To solve the given initial value problem we use the Laplace transform .Here ${u}_{p}i\left(t\right)$ is Heaviside function .
We require few results of Laplace transform and inverse Laplace transform.
$L\left\{1\right\}=\frac{1}{s}$
$L\left\{{x}^{\left(n\right)}\right\}={s}^{n}X\left(s\right)-\sum _{k=0}^{n-1}{s}^{n-1-k}{x}^{k}\left(0\right)\right)$
$L\left\{{u}_{c}\left(t\right)\right\}=\frac{{e}^{-cs}}{s}$
$L\left\{\frac{1}{{s}^{n}+1}\right\}=\frac{{t}^{n}}{\left(n!\right)}$
$L\left\{{e}^{-cs}F\left(s\right)\right\}={u}_{c}\left(t\right)f\left(t-c\right)$
Step 2
Given initial value problem
${x}^{\left(4\right)}-5x"+4x=1-{u}_{\pi }\left(t\right),x\left(0\right)={x}^{\prime }\left(0\right)=x"\left(0\right)={x}^{‴}\left(0\right)=0$
Applying Laplace transform on both sides of I.V.P. , and using the results in step 1, we get
$L\left\{{x}^{\left(4\right)}-5x"+4x\right\}=L\left\{1-{u}_{\pi }\left(t\right)\right\}$
$L\left\{{x}^{\left(4\right)}\right\}-L\left\{5x"\right\}+L\left\{4x\right\}=L\left\{1\right\}-L\left\{{u}_{\pi }\left(t\right)\right\}$
${s}^{4}X\left(s\right)-{s}^{3}x\left(0\right)-{s}^{2}{x}^{\prime }\left(0\right)-sx"\left(0\right)-{x}^{‴}\left(0\right)-5\left({s}^{2}X\left(s\right)-sx\left(0\right)-{x}^{\prime }\left(0\right)\right)+4X\left(s\right)=\frac{1}{s}-\frac{{e}^{-\pi s}}{s}$
$\left({s}^{4}-5{s}^{2}+4\right)X\left(s\right)=\frac{1}{s}-\frac{{e}^{-\pi s}}{s}$
$X\left(s\right)=\frac{1-{e}^{-\pi s}}{s\left({s}^{4}-5{s}^{2}+4\right)}$
$X\left(s\right)=\frac{1-{e}^{-\pi s}}{s\left(s+1\right)\left(s-1\right)\left(s+2\right)\left(s-2\right)}$
Applying Inverse Laplace transform to get x(t)
${L}^{-1}\left\{X\left(s\right)\right\}={L}^{-1}\left(1-{e}^{\left(}-pis\right)\right)/\left(s\left(s+1\right)\left(s-1\right)\left(s+2\right)\left(s-2\right)\right)$
$x\left(t\right)=\frac{{e}^{2t}-4{e}^{t}+6-4{e}^{-t}+{e}^{-2t}}{24}-{u}_{\pi }\left(t\right)\frac{{e}^{2\left(t-\pi \right)}-4{e}^{t-\pi }+6-4{e}^{-\left(t-\pi \right)}+{e}^{-2\left(t-\pi \right)}}{24}$