Solve by Laplace transformsy"−3y′+2y=2δ(t−1)y(0)=1y′(0)=0

Step 1The given equation is: y"−3y′+2y=2δ(t−1)Applying the Laplace transform to both sides of the equation, we get s2Y(s)−sy(0)−y′(0)−3sY(s)−y(0)+2Y(s)=2e−sStep 2Substituting the values given in the question, we get s2Y(s)−sy(0)−y′(0)−3sY(s)+y(0)+2Y(s)=2e−sor, Y(s)[s2−3s+2]−s+1=2e−sor, Y(s)=2e−s+s−1s2−3s+2or, Y(s)=2e−s+s−1s2−2s−s+2or, Y(s)=2e−s+s−1(s−1)(s−2)Step 3Now, using partial fraction method: 2e−s+s−1(s−1)(s−2)=As−1+Bs−2or, 2(1−s)+s−1(s−1)(s−2)=A(s−2)+B(s−1)(s−1)(s−2)or, 1−s(s−1)(s−2)=s(A+B)−(2A+B)(s−1)(s−2)Comparing the coefficients:A+B=−1 and 2A+B=−1Solving the above two equations for the value of A and B, we getA=0 and B=−1Step 4Substituting the value of A and B in the equation 2e−s+s−1(s−1)(s−2)=As−1+Bs−2 we get2e−s+s−1(s−1)(s−2)=−1s−2taking inverse Laplace transform both sides, we getL−1[2e−s+s−1(s−1)(s−2)]=L−1[−1s−2]

Resolved: "Solve by Laplace transformsy"−3y′+2y=2δ(t−1)y(0)=1y′(0)=0"

Harlen Pritchard

Answered question

2020-11-10

Solve by Laplace transforms

y " - 3 y^{'} + 2 y = 2 δ (t - 1)

y (0) = 1

y^{'} (0) = 0

Answer & Explanation

Bertha Stark

Skilled2020-11-11Added 96 answers

Step 1

The given equation is:

y " - 3 y^{'} + 2 y = 2 δ (t - 1)

Applying the Laplace transform to both sides of the equation, we get

s^{2} Y (s) - s y (0) - y^{'} (0) - 3 s Y (s) - y (0) + 2 Y (s) = 2 e^{- s}

Step 2

Substituting the values given in the question, we get

s^{2} Y (s) - s y (0) - y^{'} (0) - 3 s Y (s) + y (0) + 2 Y (s) = 2 e^{- s}

or, Y (s) [s^{2} - 3 s + 2] - s + 1 = 2 e^{- s}

or, Y (s) = \frac{2 e^{- s} + s - 1}{s^{2} - 3 s + 2}

or, Y (s) = \frac{2 e^{- s} + s - 1}{s^{2} - 2 s - s + 2}

or, Y (s) = \frac{2 e^{- s} + s - 1}{(s - 1) (s - 2)}

Step 3

Now, using partial fraction method:

\frac{2 e^{- s} + s - 1}{(s - 1) (s - 2)} = \frac{A}{s - 1} + \frac{B}{s - 2}

or, \frac{2 (1 - s) + s - 1}{(s - 1) (s - 2)} = \frac{A (s - 2) + B (s - 1)}{(s - 1) (s - 2)}

or, \frac{1 - s}{(s - 1) (s - 2)} = \frac{s (A + B) - (2 A + B)}{(s - 1) (s - 2)}

Comparing the coefficients:

A + B = - 1 and 2 A + B = - 1

Solving the above two equations for the value of A and B, we get

A = 0 and B = - 1

Step 4

Substituting the value of A and B in the equation \frac{2 e^{- s} + s - 1}{(s - 1) (s - 2)} = \frac{A}{s - 1} + \frac{B}{s - 2} we get

\frac{2 e^{- s} + s - 1}{(s - 1) (s - 2)} = \frac{- 1}{s - 2}

taking inverse Laplace transform both sides, we get

L^{- 1} [\frac{2 e^{- s} + s - 1}{(s - 1) (s - 2)}] = L^{- 1} [\frac{- 1}{s - 2}]

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