# y"-8y'+41y=0 y(0)=0 y'(0)=5 First , using Y for the Laplace transform of y(t), i.e., Y=Lleft{y(t)right} find the equation you get by taking the Laplace transform of the differential equation

Question
Laplace transform
$$y"-8y'+41y=0$$
$$y(0)=0$$
$$y'(0)=5$$ First , using Y for the Laplace transform of y(t), i.e., $$Y=L\left\{y(t)\right\}$$ find the equation you get by taking the Laplace transform of the differential equation

2021-02-15
Step 1
Given
$$y"-8y'+41y=0$$
$$y(0)=0$$
$$y'(0)=5$$
Step 2
Calculation
Apply Laplace transform.
$$L\left\{y"\right\}=s^2Y(s)-sy(0)-y'(0)$$
$$L\left\{y'\right\}=sY(s)-y(0)$$
$$L\left\{y\right\}=Y(s)$$
$$L\left\{c\right\}=\frac{c}{s}$$
$$L\left\{y"\right\}-8L\left\{y'\right\}+41L\left\{y\right\}=L\left\{0\right\}$$
$$\Rightarrow s^2Y(s)-sy(0)-y'(0)-8[sY(s)-y(0)]+41Y(s)=\frac{0}{s}$$
$$\Rightarrow s^2Y(s)-0-5-8[sY(s)-0]+41Y(s)=0$$
Hence the equation is
$$s^2Y(s)-0-5-8[sY(s)-0]+41Y(s)=0$$
Now,
$$s^2Y(s)-0-5-8[sY(s)-0]+41Y(s)=0$$
$$\Rightarrow Y(s)[s^2-8s+41]=5$$
$$\Rightarrow Y(s)=\frac{5}{(s^2-8s+41)}$$
Hence the value of $$Y(s)=\frac{5}{(s^2-8s+41)}$$
Apply inverse Laplace transform
$$Y(s)=\frac{5}{(s^2-8s+41)}$$
$$=\frac{5}{(s^2-8s+16+25)}$$
$$=\frac{5}{(s-4)^2+5^2}$$
$$y(t)=L^{-1}\left(\frac{5}{(s-4)^2+5^2}\right)$$
$$\Rightarrow y(t)=e^{4t} \sin(5t)$$
Hence the value of $$y(t)=e^{4t} \sin(5t)$$

### Relevant Questions

Find Y(t) using Laplace transform.
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$$y(0)=0 , y'(0)=1$$

Solution of I.V.P for harmonic oscillator with driving force is given by Inverse Laplace transform
$$y"+\omega^{2}y=\sin \gamma t , y(0)=0,y'(0)=0$$
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2) $$y(t)=L^{-1}\bigg(\frac{\gamma}{s^{2}+\omega^{2}}\bigg)$$
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4) $$y(t)=L^{-1}\bigg(\frac{\gamma}{(s^{2}+\gamma^{2})(s^{2}+\omega^{2})}\bigg)$$

Using Laplace Transform , solve the following differential equation
$${y}\text{}-{4}{y}={e}^{{-{3}{t}}},{y}{\left({0}\right)}={0},{y}'{\left({0}\right)}={2}$$
a) $$\frac{14}{{20}}{e}^{{{2}{t}}}-\frac{5}{{30}}{e}^{{-{2}{t}}}-\frac{9}{{30}}{e}^{{-{6}{t}}}$$
b) $$\frac{11}{{20}}{e}^{{{2}{t}}}-\frac{51}{{20}}{e}^{{-{2}{t}}}-\frac{4}{{20}}{e}^{{-{3}{t}}}$$
c) $$\frac{14}{{15}}{e}^{{{2}{t}}}-\frac{5}{{10}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}$$
d) $$\frac{14}{{20}}{e}^{{{2}{t}}}+\frac{5}{{20}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}$$
the differential equation be solved using the Laplace Transform? A=3
$$x^{II}(t)-3x^{I}(t)+2x(t)=A e^{2t}$$
$$x(0)=-3$$
$$x^{I}(0)=5$$
In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative $$\frac{dy}{dt}$$ also appears. Consider the following initial value problem, defined for t > 0:
$$\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}+{4}{\int_{{0}}^{{t}}}{y}{\left({t}-{w}\right)}{e}^{{-{4}{w}}}{d}{w}={3},{y}{\left({0}\right)}={0}$$
a) Use convolution and Laplace transforms to find the Laplace transform of the solution.
$${Y}{\left({s}\right)}={L}{\left\lbrace{y}{\left({t}\right)}\right)}{\rbrace}-?$$
b) Obtain the solution y(t).
y(t) - ?
Use the Laplace transform to solve the following initial value problem:
$$2y"+4y'+17y=3\cos(2t)$$
$$y(0)=y'(0)=0$$
a)take Laplace transform of both sides of the given differntial equation to create corresponding algebraic equation and then solve for $$L\left\{y(t)\right\}$$ b) Express the solution $$y(t)$$ in terms of a convolution integral
How to solve for third order differential equation of $$y"'-7y'+6y =2 \sin (t)$$ using Method of Laplace Transform when $$y(0)=0, y'(0)=0, y"(0)=0$$?
Step by step
Find the Laplace transform $$L\left\{u_3(t)(t^2-5t+6)\right\}$$
$$a) F(s)=e^{-3s}\left(\frac{2}{s^4}-\frac{5}{s^3}+\frac{6}{s^2}\right)$$
$$b) F(s)=e^{-3s}\left(\frac{2}{s^3}-\frac{5}{s^2}+\frac{6}{s}\right)$$
$$c) F(s)=e^{-3s}\frac{2+s}{s^4}$$
$$d) F(s)=e^{-3s}\frac{2+s}{s^3}$$
$$e) F(s)=e^{-3s}\frac{2-11s+30s^2}{s^3}$$
a)$$f(t)=1+2t$$ b)$$f(t) =\sin \omega t \text{Hint: Use Euler’s relationship, } \sin\omega t = \frac{e^(j\omega t)-e^(-j\omega t)}{2j}$$
c)$$f(t)=\sin(2t)+2\cos(2t)+e^{-t}\sin(2t)$$
Use the definition of Laplace Transforms to find $$L\left\{f(t)\right\}$$
$$f(t)=\begin{cases}-1 & 0\leq t <1\\1 & t\geq 1\end{cases}$$
Then rewrite f(t) as a sum of step functions, $$u_c(t)$$, and show that by taking Laplace transforms, this yields the same answer as your direct computation.