Question

Solve the initial value problem

y′′+3y=0

y(0)=1
y′(0)=1

Answer 1

The solution is
`y(t) = (4)/(3) - (1)/(3) e^{_(-3) t}\\`

Explanation:

The question is:

Solve the initial value problem

y′′+3y'=0

y(0)=1

y′(0)=1

Explanation:

First, we will write the characteristic equation, which is

`x^(2) + 3x = 0`

Then, we will solve for
`x`

`x^(2) + 3x = 0` becomes

`x(x+3) = 0\\`

∴
`x = 0` or
`x +3 = 0`

`x = 0` or
`x = -3`

Hence,
`x_(1) = 0, x_(2) = -3`

Since the two roots are distinct, that is,
`x_(1) \\eq x_(2)`

The general solution is

`y(t) = C_(1)e^{x_(1) t} + C_(2)e^{x_(2) t}\\`

Then,

`y(t) = C_(1)e^(0 t) + C_(2)e^(-3 t)`

`y(t) = C_(1) + C_(2)e^(-3 t)\\`

Then, for
`y'(t)`

`y'(t) = -3C_(2)e^(-3 t)\\`

Now, from the question

y(0)=1

y′(0)=1

Then,

`1 = y(0) = C_(1) + C_(2)e^(-3 (0))\\`

Then,

`1 = C_(1) + C_(2)` ........ (1)

Also,

`1 = y'(0) = -3C_(2)e^(-3 (0))\\`

Then,

`1 = -3C_(2)e^(-3 (0))\\`

`1 = -3C_(2)\\` ....... (2)

∴
`C_(2) = - (1)/(3)`

From equation (1)

`1 = C_(1) + C_(2)`

Then,

`C_(1) = 1 - C_(2)`

`C_(1) = 1 - -(1)/(3) \\C_(1) = 1 +(1)/(3) \\C_(1) = (4)/(3)`

Hence,
`C_(1) = (4)/(3)` and
`C_(2) = - (1)/(3)`

Then, the solution becomes

`y(t) = (4)/(3) - (1)/(3) e^{_(-3) t}\\`