Question

Use the laplace transform to solve the given initial-value problem. y'' − 8y' 16y = t, y(0) = 0, y'(0) = 1

Answer 1

If the initial value problem is $y^{11} -8y^{1} -15y=0$ and $y^{1}(0) =1$,y(0)=0 then y(t)=$(e^{3t} -e^{5t} )/2$.

Given the initial value problem be   $y^{11} -8y^{1} -15y=0$and  $y^{1}(0) =1$,y(0)=0.

We are required to find the solution of the given initial value problem.

Laplace transform is an integral transformation that converts a function of a real variable to a function of a complex variable.

Take laplace on the DE, we get

$s^{2}-sY(0)-y^{i}(0)-8[sY(s)-y(0)-15Y(s)]=0$

$s^{2}Y(s)-s(0)-1-8{sY(s)-0)}+15Y(s)=0$

(Putting the values given in question)

Y(s)=($s^{2} -8s+15)-1=0$

Y(s)=1/($s^{2} -8s+15$)

Simplifying the above:

=1/($s^{2} -5s-3s+15)$

=1/[s(s-5)-3(s-5)]

=1/2 [1/(s-3)-1/(s-5)]

Taking inverse of the above we get,

y(t)=$(e^{3t} -e^{5t} )/2$

Hence if the initial value problem is $y^{11} -8y^{1} -15y=0$ and $y^{1}(0) =1$,y(0)=0 then y(t)=$(e^{3t} -e^{5t} )/2$.

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