Question Solve the IVP dy dy +2 +26y = 0, y(0) = 0, y(0) = -5 dt2 dt = = The Laplace transform of the solutions is L{y} = = By completing the square in the denominator we see that this is the Laplace transform of shifted by the rule (Your first answer blank for this question should be a function of t). Therefore the solution is y =
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soln\frac{d^{2} y}{d t^{2}}+2 \frac{d y}{d t}+26 y=0 \quad y(0)=0, \quad y^{\prime}(0)=-5Taking laplace on both side we get\begin{array}{c}L\left\{\frac{d^{2} y}{d t^{2}}+2 \frac{d y}{d t}+2 p y\right\}=0 \\\left.L\left\{\frac{d^{2} y}{d t^{2}}\right\}+2 L\left\{\frac{d y}{d y}\right\}+26 L y\right\}=0 \\s^{2} Y(5)-s y(0)-y^{\prime}(0)+2(s Y(5)-y(0)) \neq 26 Y(s)=0 \\\left(s^{2}+2 s+26\right) Y(5)=-5 \\Y(5)=\frac{-5}{s^{2}+2 s+26}\end{array}Tuking invecse laplace on both side wegel\begin{aligned}L^{\rightarrow}\{Y(s)\}^{6} & =L^{-1}\left\{\frac{-5}{s^{2}+25+26}\right\} \\y(t) & =L^{-1}\left\{\frac{-5}{s^{2}+2 s+1+25}\right\} \\& =L^{-1}\left\{\frac{-5}{(s+1)^{2}+25}\right\} \\& =-e^{-t}\left(t^{-1}\left\{\frac{5}{s^{2}+25}\right\}\right. \text { by first shifting propedy } \\y(t) & =-e^{-t} \sin (5 t)\end{aligned}Which is required solution ...