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#### karush

##### Well-known member

- Jan 31, 2012

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$$y''-2y=e^{2t} \quad y(0)=2$$

not sure about the $e^{2t}$

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- Jan 31, 2012

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$$y''-2y=e^{2t} \quad y(0)=2$$

not sure about the $e^{2t}$

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First things first...what is the homogeneous solution?

$$y''-2y=e^{2t} \quad y(0)=2$$

not sure about the $e^{2t}$

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- Jan 31, 2012

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ok first I noticed this should be

$\tiny{2.1.17}$

\begin{align*}

\displaystyle

y'-2y&=e^{2t} \quad y(0)=2\\

u(v)&=-\exp\int 2 \, dt=-e^{2t}\\

-e^{2t}y'+2e^{2t}&=-e^{4t}

\end{align*}

so far ?

$$y'-2y=e^{2t} \quad y(0)=2$$ |

$\tiny{2.1.17}$

\begin{align*}

\displaystyle

y'-2y&=e^{2t} \quad y(0)=2\\

u(v)&=-\exp\int 2 \, dt=-e^{2t}\\

-e^{2t}y'+2e^{2t}&=-e^{4t}

\end{align*}

so far ?

Last edited:

- Admin
- #4

Okay, your integrating factor should be:ok first I noticed this should be$\tiny{2.1.17}$

$$y'-2y=e^{2t} \quad y(0)=2$$

\begin{align*}

\displaystyle

y'-2y&=e^{2t} \quad y(0)=2\\

u(v)&=-\exp\int 2 \, dt=-e^{2t}\\

-e^{2t}y'+2e^{2t}&=-e^{4t}

\end{align*}

so far ?

\(\displaystyle \mu(t)=\exp\left(-2\int\,dt\right)=e^{-2t}\)

You cannot "pull" the negative sign out of the exp() function like that, since \(\displaystyle e^{-x}\ne-e^x\)

Now, give that a go.

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- #5

- Jan 31, 2012

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I'm going to continue this and some more tomorrow

just really hard to do this on a small tablet

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- #7

- Jan 31, 2012

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What is the Characteristic Equation?When are you going to give up on the integrating factor as your only tool and use the Characteristic Equation?

$\tiny{2.1.17}$

\begin{align*}

\displaystyle

y'-2y&=e^{2t} \quad y(0)=2\\

u(v)&=\exp\int 2 \, dt=e^{-2t}\\

e^{-2t}y'+2e^{-2t}y&=1\\

(e^{-2t}y)'&=1\\

e^{-2t}y&=\int \, dt =t+c\\

y& = c_1 e^{2 t} + e^{2 t} t

\end{align*}

So if $y(0)=2$ then,

\begin{align*}

y(0)&=c e^{2(0)}+e^{2(0)}(0)=2\\

&= c(1)+0=2\\

c&=2

\end{align*}

finally

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

or

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

ok think this is it

Last edited:

- Admin
- #8

That's correct.What is the Characteristic Equation?

$\tiny{2.1.17}$

\begin{align*}

\displaystyle

y'-2y&=e^{2t} \quad y(0)=2\\

u(v)&=\exp\int 2 \, dt=e^{-2t}\\

e^{-2t}y'+2e^{-2t}y&=1\\

(e^{-2t}y)'&=1\\

e^{-2t}y&=\int \, dt =t+c\\

y& = c_1 e^{2 t} + e^{2 t} t

\end{align*}

So if $y(0)=2$ then,

\begin{align*}

y(0)&=c e^{2(0)}+e^{2(0)}(0)=2\\

&= c(1)+0=2\\

c&=2

\end{align*}

finally

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

or

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

ok think this is it

- Aug 30, 2012

- 1,207

That's correct.

@karush: The integrating factor is u(t), not u(v). A picky point but necessary.

-Dan

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- #10

Also, \(\displaystyle \exp()\) is defined as a function, and should have bracketing but I point things out once. If it's ignored then I leave it alone. I've pointed both out in the past, and that's why I was silent on them this time around.Almostall right.

@karush: The integrating factor is u(t), not u(v). A picky point but necessary.

-Dan

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- #11

- Jan 31, 2012

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that's interesting

the text book does not show the $\exp()$

surprised this got so many views

the text book does not show the $\exp()$

surprised this got so many views