Answer
$$
y=c e^{-3 t}+(t / 3)-(1 / 9)+e^{-2 t}
$$
Work Step by Step
\( \mathrm{W}^{\prime}+3 \mathrm{y}=\mathrm{t}+\mathrm{e}^{2} \)
We solve this by the integrating factor method:
\( \mu(t)=e^{k(x) d t} \)
Here, \( \mathrm{p}(\mathrm{t})=3 \)
Therefore, \( \mu(t)=e^{p_{t}} \)
\( \mu(t)=e^{x} \)
Now, multiplying both sides of the equation by \( \mu(t) \) :
\( e^{x}\left(y^{\prime}+3 y\right)=e^{x}\left(t+e^{z}\right) \)
\( e^{x y}+3 e^{x y}=e^{x} t+e^{x} e^{-x} \)
\( e^{2} y^{\prime}+3 e^{x y}=e^{x} t+e^{t} \)
Notice the left hand side of the equation:
\( e^{x} y^{\prime}+3 e^{x} y=\left(e^{x} y\right)^{\prime} \)
Therefore, the equation can be written as:
\( \left(e^{2} y\right)^{\prime}=e^{x} t+e^{i} \)
Then, integrating on both sides:
\( \int\left(e^{x} y\right)^{\prime} d t=\int\left(e^{x} t+e^{i}\right) d t \)
\( \mathrm{e}^{\mathrm{x}} \mathrm{y}=\int \mathrm{e}^{\mathrm{x}} \mathrm{d} \mathrm{d}+\int \mathrm{e} \mathrm{d} \mathrm{t} \)
On the right hand side:
\( \int e^{2} t d t=\left(t e^{x} / 3\right)-\left(e^{x} / 9\right) \)
(by using \( \int u v^{\prime}=u v-\int u^{\prime} v, \) where \( u=t \) and \( \left.v^{\prime}=e^{x}\right) \) And, \intedt = e Therefore:
(c here is the integrating constant) Dividing both sides with ex:
\( y=(t / 3)-(1 / 9)+e^{i x}+c e^{x} \)
The answer is: \( y=c e^{-s+}(t / 3)-(1 / 9)+e^{-z} \)
