Solve the following IVP using Laplace Transform y'-2y =1-t , y(0)=4

Question

Solve the following IVP using Laplace Transform

y^{'} - 2 y = 1 - t, y (0) = 4

Answer 1

Step 1
The Laplace transform is given as:

y' - 2 y = 1 - t, y (0) = 4

Taking the laplace trasform on both sides:

L {y'} - 2 L {y} = L (1) - L (t)

s L (y) - y (0) - 2 L (y) = \frac{1}{s} - \frac{1}{s^{2}}

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

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

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

Step 2
Taking the inverse Laplace transform and make into the fraction form:

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

\frac{4 s^{2} + s - 1}{s^{2} (s - 2)} = \frac{A}{s} + \frac{B}{s^{2}} + \frac{C}{(s - 2)}

4 s^{2} + s - 1 = A s (s - 2) + B (s - 2) + C s 2

4 s^{2} + s - 1 = (A + C) s^{2} + (- 2 A + B) s - 2 B

Step 3
Compare the coefficient and find out the values of A, B and C:

4 s^{2} + s - 1 = (A + C) s^{2} + (- 2 A + B) s - 2 B A + C = 4, - 2 A + B = 1 and - 2 B = - 1 \Rightarrow B = \frac{1}{2}

- 2 A + B = 1 \Rightarrow - 2 A = 1 - \frac{1}{2} \Rightarrow A = - \frac{1}{4}

A + C = 4 \Rightarrow C = 4 - (- \frac{1}{4}) \Rightarrow C = \frac{17}{4}

\frac{4 s^{2} + s - 1}{s^{2} (s - 2)} = \frac{- \frac{1}{4}}{s} + \frac{\frac{1}{2}}{s^{2}} + \frac{\frac{17}{4}}{(s - 2)}

Step 4
Now, form equation (i):

(y) = L^{- 1} {\frac{4 s^{2} + s - 1}{s^{2} (s - 2)}} = L^{- 1} {\frac{- \frac{1}{4}}{s} + \frac{\frac{1}{2}}{s^{2}} + \frac{\frac{17}{4}}{s - 2}}

y = - \frac{1}{4} L^{- 1} {\frac{1}{s}} + \frac{1}{2} L^{- 1} {\frac{1}{s^{2}}} + \frac{17}{4} L^{- 1} {\frac{1}{s - 2}}

Using L^{- 1} {\frac{1}{s}} = (1), L^{- 1} {\frac{1}{s^{2}}} = t, L^{- 1} {\frac{1}{s - a}} = e^{a t}

y = - \frac{1}{4} + \frac{t}{2} + \frac{17}{4} e^{2 t}

Want to know more about Laplace transform?