Respuesta :
Answer:
The solution is [tex]y(t) = \frac{4}{3} - \frac{1}{3} e^{_{-3} t}\\[/tex]
Step-by-step explanation:
The question is:
Solve the initial value problem
y′′+3y'=0
y(0)=1
y′(0)=1
Step-by-step explanation:
First, we will write the characteristic equation, which is
[tex]x^{2} + 3x = 0[/tex]
Then, we will solve for [tex]x[/tex]
[tex]x^{2} + 3x = 0[/tex] becomes
[tex]x(x+3) = 0\\[/tex]
∴ [tex]x = 0[/tex] or [tex]x +3 = 0[/tex]
[tex]x = 0[/tex] or [tex]x = -3[/tex]
Hence, [tex]x_{1} = 0, x_{2} = -3[/tex]
Since the two roots are distinct, that is, [tex]x_{1} \neq x_{2}[/tex]
The general solution is
[tex]y(t) = C_{1}e^{x_{1} t} + C_{2}e^{x_{2} t}\\[/tex]
Then,
[tex]y(t) = C_{1}e^{0 t} + C_{2}e^{-3 t}[/tex]
[tex]y(t) = C_{1} + C_{2}e^{-3 t}\\[/tex]
Then, for [tex]y'(t)[/tex]
[tex]y'(t) = -3C_{2}e^{-3 t}\\[/tex]
Now, from the question
y(0)=1
y′(0)=1
Then,
[tex]1 = y(0) = C_{1} + C_{2}e^{-3 (0)}\\[/tex]
Then,
[tex]1 = C_{1} + C_{2}[/tex] ........ (1)
Also,
[tex]1 = y'(0) = -3C_{2}e^{-3 (0)}\\[/tex]
Then,
[tex]1 = -3C_{2}e^{-3 (0)}\\[/tex]
[tex]1 = -3C_{2}\\[/tex] ....... (2)
∴ [tex]C_{2} = - \frac{1}{3}[/tex]
From equation (1)
[tex]1 = C_{1} + C_{2}[/tex]
Then,
[tex]C_{1} = 1 - C_{2}[/tex]
[tex]C_{1} = 1 - -\frac{1}{3} \\C_{1} = 1 +\frac{1}{3} \\C_{1} = \frac{4}{3}[/tex]
Hence, [tex]C_{1} = \frac{4}{3}[/tex] and [tex]C_{2} = - \frac{1}{3}[/tex]
Then, the solution becomes
[tex]y(t) = \frac{4}{3} - \frac{1}{3} e^{_{-3} t}\\[/tex]
