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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.

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Substitute the result in the equation and the IC. –  user26640 Mar 10 '12 at 14:43
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Click show steps button here wolframalpha.com/input/… –  Norbert Mar 10 '12 at 14:44
    
@Norbert - is there a more simple way to do it? –  Victor Mar 10 '12 at 14:47
    
By Cauchy theorem we know that your differential equation have unique solution. It is remains to check that $y(t)=2e^{-3t}+2t+1$ is desired solution. I think this is the fastest and the cheatest approach. –  Norbert Mar 10 '12 at 14:53
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If you know how to integrate $te^t$ and $e^t$ and understand the concept of Integrating Factor, you should understand my solution. (Look for Integration using Integrating Factor as search criteria, you'll find tons of material) For instance check cse.salford.ac.uk/profiles/gsmcdonald/H-Tutorials/… –  Kirthi Raman Mar 11 '12 at 0:34

3 Answers 3

up vote 2 down vote accepted

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 $2$

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

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

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If you are wondering how to integrate $te^{3t}$, you do by parts, i.e. $\int u'v = uv -\int v'u$. Assume $u'=e^{3t}$, then $\int te^{3t} = \frac{1}{3}te^{3t} - \frac{1}{3}\int e^{3t} = \frac{1}{3}te^{3t}-\frac{1}{9}e^{3t}$ –  Kirthi Raman Mar 11 '12 at 0:39

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.

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How are you able to get to the first step in your answer? –  Victor Mar 10 '12 at 15:04
    
@Victor It's like "let's assume that $y(t) = $solution$ + z(t)$". Under this assumption we get, that $z(t) = 0$ and therefore it is the only solution. I thought you were given the solution in the task, and the only thing to do was to prove that it is unique. If you want find it at first, just follow wolfram alpha as pointed out by Norbert –  dtldarek Mar 10 '12 at 16:36

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.

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please give a complete solution since i am not even knowing what the homogenous equation meaning, therefore, cannot understand what you wrote. –  Victor Mar 10 '12 at 15:25

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