MHB -2.1.17 IVP y''-2y=e^{2t} \quad y(0)=2

[karush]

\nmh{2000}
find the solution of the given initial value problem
$$y''-2y=e^{2t} \quad y(0)=2$$

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

 

[MarkFL]

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]

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 ?

 

[MarkFL]

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:

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

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

Now, give that a go. :)

 

[karush]

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]

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

 

[karush]

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

 

[MarkFL]

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]

That's correct.

Almost all right.

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

-Dan

 

[MarkFL]

Almost all right.

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

-Dan

Also, $$\exp()$$ 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]

that's interesting
the textbook does not show the $\exp()$surprised this got so many views