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[SOLVED] 2.1.17 find the solution of the given initial value problem

karush

Well-known member
Jan 31, 2012
2,928
find the solution of the given initial value problem
$$y''-2y=e^{2t} \quad y(0)=2$$

not sure about the $e^{2t}$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
find the solution of the given initial value problem
$$y''-2y=e^{2t} \quad y(0)=2$$

not sure about the $e^{2t}$
First things first...what is the homogeneous solution?
 

karush

Well-known member
Jan 31, 2012
2,928
ok first I noticed this should be
$$y'-2y=e^{2t} \quad y(0)=2$$


$\tiny{2.1.17}$
\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:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
ok first I noticed this should be
$$y'-2y=e^{2t} \quad y(0)=2$$
$\tiny{2.1.17}$
\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 ?
Okay, your integrating factor should be:

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

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. :)
 

karush

Well-known member
Jan 31, 2012
2,928
ok that's a very helpful thing to know
I'm going to continue this and some more tomorrow

just really hard to do this on a small tablet
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
When are you going to give up on the integrating factor as your only tool and use the Characteristic Equation?
 

karush

Well-known member
Jan 31, 2012
2,928
When are you going to give up on the integrating factor as your only tool and use the Characteristic Equation?
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
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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
That's correct.
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,207
That's correct.
Almost all right.

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

-Dan
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Almost all right.

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

-Dan
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.
 

karush

Well-known member
Jan 31, 2012
2,928
that's interesting
the text book does not show the $\exp()$


surprised this got so many views