The Laplace transform of the function $L\{6{e}^{-3t}-{t}^{2}+2t-8\}$

Wierzycaz
2020-10-21
Answered

The Laplace transform of the function $L\{6{e}^{-3t}-{t}^{2}+2t-8\}$

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opsadnojD

Answered 2020-10-22
Author has **95** answers

Calculation:

Consider,

$L\{6{e}^{-3t}-{t}^{2}+2t-8\}$

The objective is to determine the Laplace Transform

Consider$f,\text{}{f}_{1},\text{}\text{and}\text{}{f}_{2}\text{}\text{be the functions whose Laplace Transform exists for}\text{}s\text{}\text{}\alpha $
and c be a constant

Then, for$s\text{}\text{}\alpha $

$L\left\{{f}_{1}+{f}_{2}\right\}=L\left\{{f}_{1}\right\}+L\left\{{f}_{2}\right\}$

Also,

$L\left\{cf\right\}=cL\left\{f\right\}$

Using above properties:

$L\{6{e}^{-3t}-{t}^{2}+2t-8\}=L\{\left\{6{e}^{-3t}\right\}-L\left\{{t}^{2}\right\}+L\left\{2t\right\}-8L\left\{1\right\}$

$=6L\left\{{e}^{-3t}\right\}-L\left\{{t}^{2}\right\}-L\left\{{t}^{2}\right\}+2L\left\{t\right\}-8L\left\{1\right\}$

Since the Laplace Transform from the Laplace Transform table:

$L\{{e}^{0\cdot t}=\frac{1}{s},\mathrm{\forall}s>0$

$L\{{e}^{a}\cdot t\}=\frac{1}{s-a},\mathrm{\forall}s>a$

$L\left\{{t}^{n}\right\}=\frac{n!}{{s}^{n+1}},\mathrm{\forall}s>0,n=1,2,3,\dots $

Using above formula:

$6L\left\{{e}^{-3t}\right\}-L\left\{{t}^{2}\right\}+2L\left\{t\right\}-8L\left\{1\right\}=\frac{6}{s-(-3)}-\frac{2!}{{s}^{2+1}}+2\times \frac{1!}{{s}^{1+1}}-8\times \frac{1}{s},$

$\mathrm{\forall}s>0$

$=\frac{6}{s+3}-\frac{2}{{s}^{3}}+\frac{2}{{s}^{2}}-\frac{8}{s},\mathrm{\forall}s>0$

Laplace transforms into

$F\left(s\right)=\frac{6}{s+3}-\frac{2}{{s}^{2}}+\frac{2}{{s}^{2}}-\frac{8}{s},\mathrm{\forall}s>0$

Conclusion:

The required Laplace transform for this function is

$F\left(s\right)=\frac{6}{s+3}-\frac{2}{{s}^{2}}+\frac{2}{{s}^{2}}-\frac{8}{s},\mathrm{\forall}s>0$

Consider,

The objective is to determine the Laplace Transform

Consider

Then, for

Also,

Using above properties:

Since the Laplace Transform from the Laplace Transform table:

Using above formula:

Laplace transforms into

Conclusion:

The required Laplace transform for this function is

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I need to solve the equation below with Euler's method:

${y}^{\u2033}+\pi y{e}^{x/3}(2{y}^{\prime}\mathrm{sin}(\pi x)+\pi y\mathrm{cos}(\pi x))=\frac{y}{9}$

for the initial conditions $y(0)=1$, ${y}^{\prime}(0)=-1/3$

So I know I need to turn the problem into a system of two first order differential equations.

Therefore ${u}_{1}={y}^{\prime}$ and ${u}_{2}={y}^{\u2033}$ I can now write the system as:

${u}_{1}={y}^{\prime}\phantom{\rule{0ex}{0ex}}{u}_{2}={\displaystyle \frac{y}{9}}-\pi y{e}^{x/3}(2{u}_{1}\mathrm{sin}(\pi x)-\pi y\mathrm{cos}(\pi x))$

How do I proceed from here?

${y}^{\u2033}+\pi y{e}^{x/3}(2{y}^{\prime}\mathrm{sin}(\pi x)+\pi y\mathrm{cos}(\pi x))=\frac{y}{9}$

for the initial conditions $y(0)=1$, ${y}^{\prime}(0)=-1/3$

So I know I need to turn the problem into a system of two first order differential equations.

Therefore ${u}_{1}={y}^{\prime}$ and ${u}_{2}={y}^{\u2033}$ I can now write the system as:

${u}_{1}={y}^{\prime}\phantom{\rule{0ex}{0ex}}{u}_{2}={\displaystyle \frac{y}{9}}-\pi y{e}^{x/3}(2{u}_{1}\mathrm{sin}(\pi x)-\pi y\mathrm{cos}(\pi x))$

How do I proceed from here?