Solve System of Differential Equations with Expert Help

find Laplace transform
${\displaystyle L^{-1}\frac{5}{s^2+2s-3}}$

$F(t)=0$
${\displaystyle =\sin \left(t\right)}$
${\displaystyle 0\le t<\pi }$
${\displaystyle \pi <t<2\pi }$
Find $L(f(t))=$?

Find the inverse Laplace transform $f(t)=L^{-1}\{F(s)\}$ of the function ${\displaystyle F\left(s\right)=\frac{8e^{-4s}}{s^2+64}}$
Use h(t-c) for the Heaviside function h_c(t) if necessary.
$f(t)=L^{-1}\left\{\frac{8e^{-4s}}{s^2+64}\right\}=$

If ${\displaystyle f\left(x\right)+x^2{\left[f\left(x\right)\right]}^5=34}$ and f(1) = 2, find f '(1).

Use the accompanying tables of Laplace transforms and properties of Laplace transforms to find the Laplace tranform of the function below.
${\displaystyle {(1+e^{-4t})}^2}$

Obtain Laplace transforms for the function ${\displaystyle \sin (t-\pi )}$ step ${\displaystyle (t-\pi )}$

The Laplace transform of ${\displaystyle u(t-2)}$ is
(a) ${\displaystyle \frac{1}{s}+2}$
(b) ${\displaystyle \frac{1}{s}-2}$
(c) ${\displaystyle e^2\frac{s}{s}\left(d\right)\frac{e^{-2s}}{s}}$

Verify, using the definition of convergence of a sequence, that the following sequences converge to the proposed limit.
a) ${\displaystyle lim\frac{2n+1}{5n+4}=\frac{2}{5}}$
b) ${\displaystyle lim\frac{2n^3}{n^3+3}=0}$
c) ${\displaystyle lim\frac{\sin \left(n^2\right)}{\sqrt[3]{n}}}$

Use Laplace transform to solve the following initial value problem:
${\displaystyle y^{″}+5y=1+t,y\left(0\right)=0,y^{\prime }\left(0\right)=4}$
A)${\displaystyle \frac{7}{25}{e}^t\cos \left(2t\right)+\frac{21}{10}{e}^t\sin \left(2t\right)}$
B) ${\displaystyle \frac{7}{2}+\frac{t}{5}-\frac{t}{25}{e}^t\cos \left(2t\right)+\frac{21}{10}{e}^t\sin \left(2t\right)}$
C) ${\displaystyle \frac{7}{25}+\frac{t}{5}}$
D) ${\displaystyle \frac{7}{25}+\frac{t}{5}-\frac{7}{25}{e}^t\cos \left(2t\right)+\frac{21}{10}{e}^t\sin \left(2t\right)}$
E) non of the above
F) ${\displaystyle \frac{7}{25}+\frac{t}{5}-\frac{7}{5}{e}^t\cos \left(2t\right)+\frac{21}{10}{e}^t\sin \left(2t\right)}$

Find the inverse Laplace transform of ${\displaystyle F\left(s\right)=\frac{7s-6}{s^2-4}}$

Using Laplave Transform, evaluate the integro-differential equation
$y^{″}(x)+9y(x)=40e^x;y(0)=5,y^{\prime }(0)=-2$

Use the accompanying tables of Laplace transforms and properties of Laplace transforms to find the Laplace tranform of the function below.
${\displaystyle e^{-2t}\cos 6t+e^{5t}-1}$

If ${\displaystyle Lf\left(t\right)=F\left(s\right),\text{ then }Lf\left(4t\right)}$
a) ${\displaystyle \frac{1}{4}F\left(s\right)}$
b) ${\displaystyle F\left(\frac{s}{4}\right)}$
c) ${\displaystyle 4F\left(s\right)}$
d) ${\displaystyle \frac{1}{4}F\left(\frac{s}{4}\right)}$

L[y(x)]=Y(s) the following equation has been obtained when Laplace transformation has been applied to an initial value problem in y(x). find y(0)
${\displaystyle Y\left(s\right)=\frac{25(5-s^2)}{s(s^2+25)}}$
25
-26
-25
27
-27

find inverse Laplace transform of the given Function
${\displaystyle L^{-1}\frac{s-5}{{(s-5)}^2-9}}$

The Laplace transform of a continuous function over the interval ${\displaystyle [0,\infty )[0,\infty )}$ is defined by ${\displaystyle F\left(s\right)={\int }_0^{\mathrm{\infty }}{e}^{-sx}f\left(x\right)dx.}$
This definition is used to solve some important initial-value problems in differential equations, as discussed later. The domain of F is the set of all real numbers s such that the improper integral converges.
Find the Laplace transform F of the following functions and give the domain of F. ${\displaystyle f\left(x\right)=\cos \left(2x\right)}$

find the Laplace transform of f (t).
${\displaystyle f\left(t\right)=t^2\cos 2t}$

Let f(t) be a function on ${\displaystyle [0,\mathrm{\infty }]}$. The Laplace transform of f is the function F defined by the integral ${\displaystyle F\left(s\right)={\int }_0^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt}$ . Use this definition to determine the Laplace transform of the following function.
${\displaystyle f\left(t\right)=e^{-4t}\sin 9t}$

Find the laplace transform
${\displaystyle L[(t^2+4)e^{2t}-e^{-2t}\cos t]}$

determine the Laplace transform of f.
${\displaystyle f\left(t\right)=e^{-2t}\sin (t-\frac{\pi }{4})}$