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

• MHB
• karush
In summary, we discussed finding the solution of a given initial value problem with the equation $y''-2y=e^{2t}$ and the initial value $y(0)=2$. We first calculated the integrating factor to be $e^{-2t}$ and used it to solve the equation. We then used the initial value to determine the constant $c$ and finally arrived at the solution $y(t) = 2e^{2t} + e^{2t}t$, or $y = (t+2)e^{2t}$. We also discussed the use of the Characteristic Equation as an alternative method for solving these types of problems.
karush
Gold Member
MHB
\nmh{2000}
find the solution of the given initial value problem
$$y''-2y=e^{2t} \quad y(0)=2$$

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

Last edited:
karush said:
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?

ok first I noticed this should be
 $$y'-2y=e^{2t} \quad y(0)=2$$
$\tiny{2.1.17}$
\begin{align*}
\displaystyle
u(v)&=-\exp\int 2 \, dt=-e^{2t}\\
-e^{2t}y'+2e^{2t}&=-e^{4t}
\end{align*}

so far ?

Last edited:
karush said:
ok first I noticed this should be
 $$y'-2y=e^{2t} \quad y(0)=2$$
$\tiny{2.1.17}$
\begin{align*}
\displaystyle
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. :)

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

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

tkhunny said:
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
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:
karush said:
What is the Characteristic Equation?

$\tiny{2.1.17}$
\begin{align*}
\displaystyle
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.

MarkFL said:
That's correct.
Almost all right.

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

-Dan

topsquark said:
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.

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

## 1. What is an IVP and how does it relate to this equation?

An IVP (Initial Value Problem) is a type of differential equation that involves an initial condition, in this case y(0)=2, which is used to find the unique solution to the equation.

## 2. What is the meaning of y'' and how is it different from y'?

y'' represents the second derivative of the function y, which measures the rate of change of the rate of change of y. This is different from y', which represents the first derivative of y and measures the rate of change of y.

## 3. What does the term e^{2t} represent in this equation?

e^{2t} is an exponential function with a base of e (Euler's number) raised to the power of 2t. In this equation, it is the input or forcing function that is causing the change in y.

## 4. How is this equation solved and what is the solution?

This equation can be solved using the method of undetermined coefficients or variation of parameters. The solution to this specific equation is y(t) = 2e^{2t} + \frac{3}{4}e^{-t}.

## 5. What is the significance of the initial condition y(0)=2 in this equation?

The initial condition y(0)=2 serves as a starting point or boundary condition for the solution of the equation. It allows us to find the specific solution that satisfies the given condition.

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