Solve the IVP with Laplace Transform:{y"+4y′+4y=(3+t)e−2ty(0)=2y′(0)=5

Step 1The equation is:y"+4y′+4y=(3+t)e−2tNow Laplace equation will be:[s2L{y}−sy(0)−y′(0)]+4[sL{y}−y(0)]+4L{y}=31s+2+1(s+2)2(s2+4s+4)L{y}−2s−5−8=3s+2+1(s+2)2(s2+4s+4)L{y}−2s−13=3s+2+1(s+2)2(s2+4s+4)L{y}=2s+13+3s+2+1(s+2)2Step 2L{y}=2s3+21s2+63s+59(s+2)2(s)+4s+4L{y}=2s3+21s2+63s+59(s+2)4Now take Laplace inverse,y=L−1{2s3+21s2+63s+59(s+2)4}By the formula of Laplace transformation the final solution will be:y=(2+9x)e−2x

Find out the answer to "Solve the IVP with Laplace Transform:{y"+4y′+4y=(3+t)e−2ty(0)=2y′(0)=5"

tinfoQ

Answered question

2020-12-21

Solve the IVP with Laplace Transform:

{\begin{cases} y " + 4 y^{'} + 4 y = (3 + t) e^{- 2 t} \\ y (0) = 2 \\ y^{'} (0) = 5 \end{cases}

Answer & Explanation

Sadie Eaton

Skilled2020-12-22Added 104 answers

Step 1

The equation is:

y " + 4 y^{'} + 4 y = (3 + t) e^{- 2 t}

Now Laplace equation will be:

[s^{2} L {y} - s y (0) - y^{'} (0)] + 4 [s L {y} - y (0)] + 4 L {y} = 3 \frac{1}{s + 2} + \frac{1}{(s + 2)^{2}}

(s^{2} + 4 s + 4) L {y} - 2 s - 5 - 8 = \frac{3}{s + 2} + \frac{1}{(s + 2)^{2}}

(s^{2} + 4 s + 4) L {y} - 2 s - 13 = \frac{3}{s + 2} + \frac{1}{(s + 2)^{2}}

(s^{2} + 4 s + 4) L {y} = 2 s + 13 + \frac{3}{s + 2} + \frac{1}{(s + 2)^{2}}

Step 2

L {y} = \frac{2 s^{3} + 21 s^{2} + 63 s + 59}{(s + 2)^{2} (s)} + 4 s + 4

L {y} = \frac{2 s^{3} + 21 s^{2} + 63 s + 59}{(s + 2)^{4}}

Now take Laplace inverse,

y = L^{- 1} {\frac{2 s^{3} + 21 s^{2} + 63 s + 59}{(s + 2)^{4}}}

By the formula of Laplace transformation the final solution will be:

y = (2 + 9 x) e^{- 2 x}

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