Question

Solve the differential equation using laplace transform:

Y" – 7y' = 6e31 – 3e? y(0) = 1, y'(O) = (-1)

Answer 1

$\text {First by using the operator for Laplace transform for the Differential equation.} \\ L(y'') = s^2y(s) - sy(0) - y'(0) \\ L(y') = sy(s) - y(0) \\ L(y) = y(s) \\ \text { Some more useful Laplace operations are, } \\ L ( e^{at}) = \frac{1}{s-a} \\ L(1)=\frac{1}{s}$

and for inverse,

L-1 sa) 1 S-a = (eat) 1-7 = 1 S

3e+ Write down the question below, y' – 7y' = 6e3 Take the laplace on both side, = L(' – 7y') = L(6e34 – 3e) =L(') - 7L(y') = L(63) - L(3e) By using the respective laplace operators, → (sạy(s) – sy(0) – y'(0) – 7(sy(s) – y(0)) = 6 (+)-3(5+1)

-12 -3 1 S +1 S- 3 → (sạy(s) – sy(0) - y'(0) – 7(sy(s) – y(0)) = 6 Apply the boundary conditions, = (s’y(s) – s(1)-(-1)) – 7(sy(s) – (1)) -3 3(3+1) S — → s’y(s) – 5+1 – 7sy(s) + 7 = 6 -3 S S+ ► sy(s) - 5+8 – 7sy(s) = 6 - 3 S- 3 S+1 → sʻy(s) – 7sy(s) = 5-8+6 ( -3 S-3 S +1

sy(s) - 7sy(s) = s - 8+6 - 3 S 3 S +1 1 =>y(s) = S-8 + 6 3 1)) (52 – 7s) S- 3 s +1 1 1 1 y(s) S s(s – 7) 1 8 +6 - 3 s(s – 7) s(s – 7)(s – 3) s(s – 7)(8 +1) 1 +6 s(s – 7) - 3 s(s – 7)(s - 3) s(s – 7)(8 +1) = y(s) = 8 3 (818 (s-7)

$\\ \text { Now, take the inverse of Laplace, } \\ \Rightarrow L^{-1}(y(s)) = L^{-1}\left (\frac{1}{(s - 7)} - 8\frac{1}{s(s - 7)} + 6\left ( \frac{1}{s(s - 7)(s-3)} \right ) - 3\left ( \frac{1}{s(s - 7)(s+1)} \right ) \right )\\ \text { First we have to perform the partial fraction of respective fractions, } \\$

$\\ \Rightarrow L^{-1}(y(s)) = L^{-1}\left (\frac{1}{(s - 7)} - \frac{8}{s(s - 7)} + \left ( \frac{6}{s(s - 7)(s-3)} \right ) - \left ( \frac{3}{s(s - 7)(s+1)} \right ) \right )\\ \text { First find the partial fraction of, } \left (\frac{8}{s(s - 7)} \right ) \\ \Rightarrow \left (\frac{8}{s(s - 7)} \right ) = \frac{A}{s} + \frac{B}{s-7} \\ \text { Take LCM on right hand side, }\\ \Rightarrow \left (\frac{8}{s(s - 7)} \right ) = \frac{A(s-7)}{s(s-7)} + \frac{Bs}{s(s-7)} \\ \Rightarrow 8 = A(s-7) + Bs \\ \Rightarrow 8 = As-7A + Bs \\ \Rightarrow 8 = (A+B)s-7A \\ \text { Compare the coefficient on boths side, } \\ (A+B) = 0 \\ -7A = 8 \\ A = \frac{-8}{7} \\ B = \frac{8}{7} \\ \Rightarrow \left (\frac{8}{s(s - 7)} \right ) = -\frac{8}{7s} + \frac{8}{7(s-7)}$

$\text { Now, find the partial fraction of, } \left ( \frac{6}{s(s - 7)(s-3)} \right ) \\ \Rightarrow \left ( \frac{6}{s(s - 7)(s-3)} \right ) = \frac{A}{s} + \frac{B}{(s-7)} + \frac{C}{(s-3)} \\ \Rightarrow \left ( \frac{6}{s(s - 7)(s-3)} \right ) = \frac{A(s - 7)(s-3)}{s(s - 7)(s-3)} + \frac{Bs(s-3)}{s(s - 7)(s-3)}+ \frac{Cs(s - 7)}{s(s - 7)(s-3)} \\ \Rightarrow 6 = A(s - 7)(s-3) + {Bs(s-3)} + {Cs(s - 7)} \\ \text { Expand and compare the coefficient, } \\ \text { we get the value of } A, B \text { and } C\\ A = \frac{2}{7} \\ B = \frac{3}{14} \\ C = \frac{-1}{2} \\ \text { The above partial fraction can be written as, } \\ \Rightarrow \left ( \frac{6}{s(s - 7)(s-3)} \right ) = \frac{2}{7s} + \frac{3}{14(s-7)} - \frac{1}{2(s-3)} \\$

$\text { Follow the same process and find the partial fraction of } \frac{3}{s(s - 7)(s+1)}, \\ \Rightarrow \frac{3}{s(s - 7)(s+1)} \\ \text { The above fraction can be written as, } \\ \Rightarrow \frac{3}{s(s - 7)(s+1)} = -\frac{3}{7s}+\frac{3}{56\left(s-7\right)}+\frac{3}{8\left(s+1\right)}\\$

Substitute the following relation in the below equation,

$\Rightarrow L^{-1}(y(s)) = L^{-1}\left (\frac{1}{(s - 7)} - \frac{8}{s(s - 7)} + \left ( \frac{6}{s(s - 7)(s-3)} \right ) - \left ( \frac{3}{s(s - 7)(s+1)} \right ) \right )\\ \Rightarrow L^{-1}(y(s)) = L^{-1}\left (\frac{1}{(s - 7)} + \frac{8}{7s}-\frac{8}{7(s-7)} + \frac{2}{7s} + \frac{3}{14(s-7)} - \frac{1}{2(s-3)} + \frac{3}{7s}-\frac{3}{56\left(s-7\right)}-\frac{3}{8\left(s+1\right)} \right )\\ \Rightarrow L^{-1}(y(s)) = L^{-1}\left (\frac{1}{56(s - 7)} + \frac{13}{7s} - \frac{1}{2(s-3)} -\frac{3}{8\left(s+1\right)} \right )\\$

now by using Laplace inverse operator,

we get.

$y(t)=\frac{13}{7}-\frac{3}{8}e^{-t}-\frac{1}{2}e^{3t}+\frac{1}{56}e^{7t}$