First Order Ordinary Differential Equation?!

(dy/dt) + P(t)y = t, y(0) = 1,

With P(t) = {1, 0 <_ t <_ 2,
3, t > 2.

a) Just wondering how one would find the general solution for 0 <_ to t <_ to 2?

b) Also how to find the constant in the general solution for 0 <_ t <_ 2 so that the initial condition is satisfied.

(c) Find a general solution for t > 2

(d) Find the constant in the general solution for t > 2 so that the solution for t > 2
and the solution for 0 <_ t <_ 2 match at t = 2

Thank you so much to anyone who answers any of these!

y' + P(t)y = t
P(t) = 1
y' + y = t
````````````integrating factor = e^(∫1 dt) = eᵗ
eᵗy' + eᵗy = teᵗ
(eᵗy)' = teᵗ

∫(eᵗy)' = ∫ teᵗ dt
eᵗy = eᵗ(t - 1) + C
y = t - 1 + Ce⁻ᵗ
``````````````````
b)
y(0) = 1
y = t - 1 + Ce⁻ᵗ
1 = 0 - 1 + C
C = 2
y = t - 1 + 2e⁻ᵗ
``````````````````
c)
y' + P(t)y = t
P(t) = 3
y' + 3y = t
````````````integrating factor = e^(∫3 dt) = e³ᵗ
e³ᵗy' + e³ᵗy = t e³ᵗ
(e³ᵗy)' = t e³ᵗ
y = (3t - 1)/9 + Ce⁻³ᵗ

d)
0 ≤ t ≤ 2
y = t - 1 + 2e⁻ᵗ
t = 2,
y = 2 - 1 + 2e⁻²
y = 1 + 2/e²

t > 2
y = (3t - 1)/9 + Ce⁻³ᵗ
1 + 2/e² = (3*2 - 1)/9 + Ce⁻⁶
(e² + 2)/e² = 5/9 + C/e⁶
C = 2e⁴ + 4e⁶ / 9
```````````````````````≈ 288.5