Solving an ordinary differential equation: $y'(t)+3y(t)=6t+5$

Question

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.

Answer 1

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 $2$

$$ e^{3t}y(t) = 2t e^{3t} +e^{3t} + 2$$

$$ \Rightarrow y(t) = 2t + 1 + 2e^{-3t}$$

Answer 2

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.

Answer 3

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.