# Find the solution of the following differential equation by Laplace transforms: y'''- 5y" + 7y’-3y =20sin(t) , y(0)=y'(0)=0 , y"(0)=-2

Question
Laplace transform
Find the solution of the following differential equation by Laplace transforms:
$$y'''- 5y" + 7y’-3y =20\sin(t) , y(0)=y'(0)=0 , y"(0)=-2$$

2020-12-16
Step 1
Consider the following initial value problem and apply the Laplace transform on both sides:
$$y'''-5y"+7y'-3y=20\sin t$$
$$y(0)=y'(0)=0 , y"(0)=-2$$
$$s^3L\left\{y(t)\right\}-s^2y(0)-sy'(0)-y"(0)-5\left\{s^2L\left\{y(t)\right\}-sy(0)-y'(0)\right\}+7\left\{sL\left\{y(t)\right\}-y(0)\right\}-3L\left\{y(t)\right\}=\frac{20}{(s^2+1)}$$
$$s^3Y\left\{y(t)\right\}+2-5\left\{s^2L\left\{y(t)\right\}\right\}+7\left\{sL\left\{y(t)\right\}\right\}-3L\left\{y(t)\right\}=\frac{20}{(s^2+1)}$$
$$(s^3-5s^2+7s-3)L\left\{y(t)\right\}+2=\frac{20}{(s^2+1)}$$
$$(s^3-5s^2+7s-3)L\left\{y(t)\right\}=\frac{20}{(s^2+1)}-2$$
$$L\left\{y(t)\right\}=\frac{20}{(s^2+1)(s^3-5s^2+7s-3)}-\frac{2}{(s^3-5s^2+7s-3)}$$
$$=\frac{1-3s}{(s^2+1)}+\frac{3}{(s+1)}-\frac{4}{(s-1)^2}$$
$$=\frac{1}{(s^2+1)}+\frac{3}{(s+1)}-\frac{4}{(s-1)^2}-\frac{3s}{(s^2+1)}$$
Step 2
Apply the inverse Laplace transform:
$$L\left\{y(t)\right\}=\frac{1}{(s^2+1)}+\frac{3}{(s+1)}-\frac{4}{(s-1)^2}-\frac{3s}{(s^2+1)}$$
$$\left\{y(t)\right\}=L^{-1}\left[\frac{1}{(s^2+1)}+\frac{3}{(s+1)}-\frac{4}{(s-1)^2}-\frac{3s}{(s^2+1)}\right]$$
$$y(t)=\sin t +3e^t-4te^t-3\cos t$$

### Relevant Questions

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) - ?
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
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Use the Laplace transform to solve the given differential equation subject to the indicated initial conditions.
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$$y(0) = 0$$
$$y'(0) = 0$$
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$$y(0,t)=0$$
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$$\lim_{x\rightarrow\infty}y_x(x.t)=0$$
$$y"-8y'+41y=0$$
$$y(0)=0$$
$$y'(0)=5$$ 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
$$y"-3y'-4y=3e^{2x}$$
$$y(0)=1 , y'(0)=0$$
$$y"-3y'+2y=2\delta(t-1)$$
$$y(0)=1$$
$$y'(0)=0$$