Prove if y'(t)+3y(t)=6t+5, $y(0)=3$, then $y(t)=2e^{-3t}+2t+1$.
I have no idea how to finish this problem.
Prove if y'(t)+3y(t)=6t+5, $y(0)=3$, then $y(t)=2e^{-3t}+2t+1$.
I have no idea how to finish this problem.
First solve the homogeneous equation y'+3y=0. This gives a homegeneous solution $y_h$. It remains find to a particular solution $y_p$. The solution to the ODE is then given by $y=y_h+y_p$. Since the non-homogeneous part of the equation is a polynomial of degree 1, try $y_p=t+c$, where $c$ is some constant. This constant is uniquely determined by the initial condition.
Set $y(t) = 2e^{-3t}+2t+1 + z(t)$ and substitute it to the equations: y'(t)+3y(t)=6t+5 $ y(0) = 3 $ We get: -6e^{-3t}+2+z'(t)+6e^{-3t}+6t+3 + 3z(t) = 6t+5 $ 3 + z(0) = 3$ what simplifies to: z'(t)+ 3z(t) = 0 $ z(0) = 0\,, $ but this simple ODE has only one solution, namely $z(t) = 0$, and that completes the proof.
The given equation is a first order differential equation
y'(t)+3y(t)=6t+5 can be solved by finding something called an integrating factor.
For a first order differential equation of the form
$ \frac{dy}{dt} + P(t) y = Q(t)$ The integrating factor is $e^{f(t)}$ where
$f(t) = \int P(t) {\text dt}$
In this case, the integrating factor is $e^{3t}$ because
$ \int 3 {\text dt} = 3t $
Now multiply the given equation throughout by $e^{3t}$ to get
e^{3t} y'(t)+3 e^{3t}y(t)=e^{3t}\left(6t+5\right)
The left hand side is
$ \frac{d}{dt} \left(e^{3t}y\right)$
Therefore
$ \frac{d}{dt} \left(e^{3t}y\right) = e^{3t}\left(6t+5\right)$
Integrate both sides now
$ \left(e^{3t}y\right) = \int 6te^{3t} {\text dt} + 5\int e^{3t}{\text dt}$
$ \begin{align*} e^{3t}y &= 6t \frac{e^{3t}}{3} - 6 \int \frac{e^{3t}}{3} {\text dt} + \frac{5}{3} e^{3t}\\ &= 6t \frac{e^{3t}}{3} +e^{3t} + {\text constant}\\ &= 2t e^{3t} +e^{3t} + {\text constant} \end{align*} $ But $y(0) = 3$ therefore the constant factor is $2$
$ e^{3t}y(t) = 2t e^{3t} +e^{3t} + 2$
$ \Rightarrow y(t) = 2t + 1 + 2e^{-3t}$