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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.2012-03-10

3 Answers 3

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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.2012-03-10
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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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    @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](http://www.wolframalpha.com/input/?i=y%27%5Bt%5D%2B3y%5Bt%5D%3D6t%2B5%2C+y%5B0%5D%3D3) as pointed out by Norbert2012-03-10
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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}$

  • 0
    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}$2012-03-11