# How to solve this problem? Use the Laplace transform to solve the given initial value problem. y'' - y' -6y = 0; y(0)=1, y'(0)=-1answer is y=(1/5)(e^3t + 4e^-2t)

hala718 | High School Teacher | (Level 1) Educator Emeritus

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`y'' - y' - 6y = 0`

`==gt L {y'' - y' - 6y} = 0`

`==gt L{y''}(s) - L{y'}(s) - 6{y}(s) = 0` ...........(1)

Let `y(s)= L{y}(s) `

`==gt L{y'}(s) = sy(s) - y(0)= sy(s) -1`

` ==gt L{y''}(s)= s^2 y(s)- sy(0) - y'(0)= s^2 y(s) -s +1`

Now we will substitute into (1)

`==gt s^2 y(s) -s +1 - (sy(s)-1) - 6y(s)= 0 `

`==gt s^2 y(s) - sy(s) - 6y(s) -s+1 +1 = 0 `

`==gt y(s) (s^2 -s -6) = s- 2 `

`==gt y(s)= (s-2)/(s^2-s-6) `

`==gt y(s)= (s-2)/((s-3)(s+2))`

Now we will rewrite in partial fractions.

`==gt (s-2)/((s-3)(s+2)) = A/(s-3) + B/(s+2)`

==> s-2 = A(s+2) + B(s-3)

==> s = (A+b)s ==> A+B = 1==> A= 1-B

==> -2 = (2A -3B)

==> -2 = 2(1-B) - 3B

==> -2 = 2-2B-3B

==> -4 = -5B

==> B= 4/5

==> A = 1/5

`==gt y(s)= (1/5)/(s-3) + (4/5)/(s+2) `

`==gt y(s)= (1/5) (1/(s-3)) + (4/5) (1/(s-(-2)) `

`==gt y(t)= (1/5) e^3t + (4/5) e^(-2t)) `

`==gt y(t) = (1/5) (e^3t + 4e^(-2t))`