## Related questions with answers

Solve the IVPs by the Laplace transform. If necessary, use partial fraction expansion as in Example 4 of the text. Show all details. y''-y'-6y=0, y(0)=11, y'(0)=28

Solution

Verified$y''-y'-6y=0\;\;\;\;,\;\;\;\;y(0)=11\;\;\;\;,\;\;\;\;y'(0)=28$

$\textbf{Note That : }\;$

$L\left\{y \right\}={\color{#19804f} {Y}}$

$L\left\{y' \right\}={\color{#4257b2} {SY-y(0)}}$

$L\left\{y'' \right\}={\color{#c34632} {S^{2}Y-Sy(0)-y'(0)}}$

$\therefore \;\; {\color{#c34632} {(S^{2}Y-11S-28)}}-{\color{#4257b2} {(SY-11)}}-{\color{#19804f} {(6Y)}}=0$

$(S^{2}+S+6)Y-11S-17=0$

$Y=\dfrac{11S+17}{S^{2}-S-6}=\boxed{\;\dfrac{11S+17}{(S-3)(S+2)}\;}$

$\textbf{By Using Partial Fraction}$

$Y=\boxed{\;\dfrac{1}{S+2}+\dfrac{10}{S-3}\;}$

$\textbf{Taking Laplace Transform on both sides : }\;$

$Where\;:\;\boxed{\;\color{#19804f} {L\left\{\dfrac{a_{0}}{S+a} \right\}=a_{0}\;e^{-at}}\;}$

$\therefore \;\;\; \textbf{Solution is : }\;\; \boxed{\color{#c34632} {y=e^{-2t}+10\;e^{3t}}\;}$

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