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Use the Laplace transform to solve the following initial value problem: y" + 8y' = 0 y(0) = -3, y'(0) = -4 First, using Y for the Laplace transform of y(t), i.e., Y = L{y(t)}, find the equation you get by taking the Laplace transform of the differential equation = 0 Now solve for Y(s) = and write the above answer in its partial fraction decomposition, Y(s) = Y(s) = A + on where a

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Answer:

Step-by-step explanation:

[tex]y" + 8y' = 0[/tex] is the given second order differential equation.

Initial conditions are [tex]y(0) = -3, y'(0) = -4[/tex]

Take Laplace for the DE we get

[tex]L(y")+8L(y')=0\\s^2Y(s)-sy(0)-y'(0) +8(sY(s)-y(0) )=0\\Y(s) (s^2+8s) +3s+4+24=0\\[/tex]

[tex]Y(s)(s^2+8s) = -3s-28\\Y(s) = \frac{-3s-28}{s^2+8s} [/tex]

Right side = [tex]\frac{-7}{2s} +\frac{1}{2(s+8)}[/tex]

Now take inverse laplace

[tex]y(t) = \frac{-7}{2} +\frac{1}{2} e^{-8t}[/tex]

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