solve the initial value problem y' +3y =f(t) ,y(0)=1 where f(t) =1 for 0&lt;=t&lt;1 and 0 t&gt;=1

Name: solve the initial value problem y&#039; +3y =f(t) ,y(0)=1 where f(t) =1 for 0&amp;lt;=t&amp;lt;1 and 0 t&amp;gt;=1
Uploaded: 2022-02-23T17:22:57+00:00
Description: solve the initial value problem y&#039; +3y =f(t) ,y(0)=1 where f(t) =1 for 0&amp;lt;=t&amp;lt;1 and 0 t&amp;gt;=1

Question

solve the initial value problem y&#039; +3y =f(t) ,y(0)=1 where f(t) =1 for 0&amp;lt;=t&amp;lt;1 and 0 t&amp;gt;=1

alenahelenash · Accepted Answer

To solve the initial value problem, we will use an integrating factor. First, we find the integrating factor by multiplying both sides of the differential equation by the exponential of the coefficient of y, which is 3. This gives us:
$e^{3 t} \cdot y' + 3 e^{3 t} \cdot y = e^{3 t} \cdot f (t)$

Notice that the left-hand side can be simplified using the product rule for derivatives, as follows:
$(e^{3 t} \cdot y)' = e^{3 t} \cdot y' + 3 e^{3 t} \cdot y$

Substituting this into our original equation, we get:
$(e^{3 t} \cdot y)' = e^{3 t} \cdot f (t)$

We can integrate both sides with respect to t to get:
$e^{3 t} \cdot y = \int [0, t] e^{3 s} \cdot f (s) d s + C$

where C is a constant of integration. To find C, we use the initial condition $y (0) = 1$ :
$e^{0} \cdot 1 = \int [0, 0] e^{3 s} \cdot f (s) d s + C$

Simplifying, we get:
$1 = C$

Substituting this back into our equation, we get:
$e^{3 t} \cdot y = \int [0, t] e^{3 s} \cdot f (s) d s + 1$

Now we just need to evaluate the integral on the right-hand side for the two cases of f(t):

Case 1: $0 \leq t < 1$
$\int [0, t] e^{3 s} \cdot f (s) d s = \int [0, t] e^{3 s} d s = (\frac{1}{3}) \cdot (e^{3 t} - 1)$

Substituting into our equation, we get:
$e^{3 t} \cdot y = (\frac{1}{3}) \cdot (e^{3 t} - 1) + 1$
Simplifying, we get:
$y = (\frac{1}{3}) \cdot \frac{e^{3 t} - 1}{e^{3 t}} + \frac{1}{e^{3 t}} = (\frac{1}{3}) - (\frac{2}{3}) \cdot e^{- 3 t}$

Case 2: $t > = 1$
$\int [0, t] e^{3 s} \cdot f (s) d s = \int [0, 1] e^{3 s} d s + \int [1, t] e^{3 s} \cdot 0 d s = (\frac{1}{3}) \cdot (e^{3} - 1)$

Substituting into our equation, we get:
$e^{3 t} \cdot y = (\frac{1}{3}) \cdot (e^{3} - 1) + 1$

Simplifying, we get:
$y = \frac{(\frac{1}{3}) \cdot (e^{3} - 1) + 1}{e^{3 t}} = (\frac{1}{3}) \cdot e^{- 3 t} + \frac{1 - \frac{1}{3} \cdot e^{- 3}}{e^{3 t}}$

Therefore, the solution to the initial value problem is:
$y = {(\frac{1}{3}) - (\frac{2}{3}) \cdot e^{- 3 t} for 0$

solve the initial value problem y' +3y =f(t)

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