Solve the following initial value problems using Laplace Transforms:a)d2ydt2+4dydt+3y=1,y(0=0),y′(0)=0b)d2ydt2+4dydt=1,y(0=0),y′(0)=0c)2d2ydt2+3dydt−2y=te−2t,y(0=0),y′(0)=−2

Name: Solve the following initial value problems using Laplace Transforms:a)d2ydt2+4dydt+3y=1,y(0=0),y′(0)=0b)d2ydt2+4dydt=1,y(0=0),y′(0)=0c)2d2ydt2+3dydt−2y=te−2t,y(0=0),y′(0)=−2
Uploaded: 2022-02-23T17:22:57+00:00
Description: Solve the following initial value problems using Laplace Transforms:a)d2ydt2+4dydt+3y=1,y(0=0),y′(0)=0b)d2ydt2+4dydt=1,y(0=0),y′(0)=0c)2d2ydt2+3dydt−2y=te−2t,y(0=0),y′(0)=−2

Step 1 We need to solve the given differential equations using Laplace transformation.a)d2ydt2+4dydt+3y=1,y(0=0),y′(0)=0Now , taking Laplace transformation both of the sides.L{d2ydt2+4dydt+3y}=L{1}⇒s2Y(s)−sy(0)−y′(0)+4(sY(s)−y(0))+3Y(s)=1s⇒(s2+4s+3)Y(s)=1s( Using y(0)=0,y′(0)=0)⇒Y(s)=1s2+4s+3⇒Y(s)=1s(s+1)(s+3)⇒Y(s)=13s−12(s+1)+16(s+3) (By the method of partial fraction)Step 2Now, taking inverse Laplace transformation to get the final answer.⇒y(t)=L−1{13s−12(s+1)+16(s+3)}⇒y(t)=13L−1{1s}−12L−1{1s+1}+16L−1{1s+3}⇒y(t)=13⋅1−12⋅e−t+16⋅e−3t⇒y(t)=13−e−t2+e−3t6Answer y(t)=13−e−t2+e−3t6Step 3b)d2ydt2+4dydt=1,y(0=0),y′(0)=0Now , taking Laplace transformation both of the sides. L{d2ydt2+4dydt}=L{1}⇒s2Y(s)−sy(0)−y′(0)+4(sY(s)−y(0))=1s⇒(s2+4s)Y(s)=1s(Using y(0)=0,y′(0)=0)⇒Y(s)=1s(s2+4s)⇒Y(s)=1s(s+1)(s+3)⇒Y(s)=14s2+116(s+4)−116s (By the method of partial fraction)Step 4Now, taking inverse Laplace transformation to get the final answer.

Step by step solution to "Solve the following initial value problems using Laplace..."

sanuluy

Answered question

2021-01-10

Solve the following initial value problems using Laplace Transforms:

a) \frac{d^{2} y}{d t^{2}} + 4 \frac{d y}{d t} + 3 y = 1,

y (0 = 0), y^{'} (0) = 0

b) \frac{d^{2} y}{d t^{2}} + 4 \frac{d y}{d t} = 1,

y (0 = 0), y^{'} (0) = 0

c) 2 \frac{d^{2} y}{d t^{2}} + 3 \frac{d y}{d t} - 2 y = t e^{- 2 t},

y (0 = 0), y^{'} (0) = - 2

Answer & Explanation

Fatema Sutton

Skilled2021-01-11Added 88 answers

Step 1 We need to solve the given differential equations using Laplace transformation.
$a) \frac{d^{2} y}{d t^{2}} + 4 \frac{d y}{d t} + 3 y = 1,$
$y (0 = 0), y^{'} (0) = 0$
Now , taking Laplace transformation both of the sides.
$L {\frac{d^{2} y}{d t^{2}} + 4 \frac{d y}{d t} + 3 y} = L {1}$
$\Rightarrow s^{2} Y (s) - s y (0) - y^{'} (0) + 4 (s Y (s) - y (0)) + 3 Y (s) = \frac{1}{s}$
$\Rightarrow (s^{2} + 4 s + 3) Y (s) = \frac{1}{s} (Using y (0) = 0, y^{'} (0) = 0)$
$\Rightarrow Y (s) = \frac{1}{s^{2} + 4 s + 3}$
$\Rightarrow Y (s) = \frac{1}{s (s + 1) (s + 3)}$
$\Rightarrow Y (s) = \frac{1}{3 s} - \frac{1}{2 (s + 1)} + \frac{1}{6 (s + 3)}$ (By the method of partial fraction)
Step 2
Now, taking inverse Laplace transformation to get the final answer.
$\Rightarrow y (t) = L^{- 1} {\frac{1}{3 s} - \frac{1}{2 (s + 1)} + \frac{1}{6 (s + 3)}}$
$\Rightarrow y (t) = \frac{1}{3} L^{- 1} {\frac{1}{s}} - \frac{1}{2} L^{- 1} {\frac{1}{s + 1}} + \frac{1}{6} L^{- 1} {\frac{1}{s + 3}}$
$\Rightarrow y (t) = \frac{1}{3} \cdot 1 - \frac{1}{2} \cdot e^{- t} + \frac{1}{6} \cdot e^{- 3 t}$
$\Rightarrow y (t) = \frac{1}{3} - \frac{e^{- t}}{2} + \frac{e^{- 3 t}}{6}$
Answer $y (t) = \frac{1}{3} - \frac{e^{- t}}{2} + \frac{e^{- 3 t}}{6}$
Step 3
$b) \frac{d^{2} y}{d t^{2}} + 4 \frac{d y}{d t} = 1,$
$y (0 = 0), y^{'} (0) = 0$
Now , taking Laplace transformation both of the sides. $L {\frac{d^{2} y}{d t^{2}} + 4 \frac{d y}{d t}} = L {1}$
$\Rightarrow s^{2} Y (s) - s y (0) - y^{'} (0) + 4 (s Y (s) - y (0)) = \frac{1}{s}$
$\Rightarrow (s^{2} + 4 s) Y (s) = \frac{1}{s} (Using y (0) = 0, y^{'} (0) = 0)$
$\Rightarrow Y (s) = \frac{1}{s (s^{2} + 4 s)}$
$\Rightarrow Y (s) = \frac{1}{s (s + 1) (s + 3)}$
$\Rightarrow Y (s) = \frac{1}{4 s^{2}} + \frac{1}{16 (s + 4)} - \frac{1}{16 s}$ (By the method of partial fraction)
Step 4
Now, taking inverse Laplace transformation to get the final answer.
$<$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Differential Equations

Didn't find what you were looking for?