-b.1.2.2c initial value problem

[karush]

$\displaystyle \frac{dy}{dt}=2y-5, \quad y(0)=y_0$
rewrite
$$y'-2y=-5$$
obtain u(x)
$$u(x)=\exp\int-2\, dx = e^{-2t}$$
then
$$(e^{-2t}y')=5e^{-2t}$$
just reviewing but kinda

? ​

 

[MarkFL]

In your last line, on the LHS, you want:

\(\displaystyle \left(e^{-2t}y\right)'\)

Or what I would write:

\(\displaystyle \frac{d}{dt}\left(e^{-2t}y\right)\)

And on the RHS, you want:

\(\displaystyle -5e^{-2t}\)

 

[karush]

$\quad\displaystyle \frac{dy}{dt}=2y-5, \quad y(0)=y_0$
rewrite
$\quad y'-2y=-5$
obtain u(x)
$\quad u(x)=\exp\int-2\, dx = e^{-2t}$
then
$\quad (e^{-2t}y)'=-5e^{-2t}$
integrate both sides
$\quad\displaystyle e^{-2t}y=-5\int e^{-2t} dt=-5e^{-2t}+c_1$
finally
$\quad\displaystyle y
=-5\frac{e^{-2t}}{e^{-2t}}+\frac{c_1}{e^{-2t}}=c_1e^{2t}-5$
so at $\quad y(0)=y_0$

ok I don't see how they got this answer

View attachment 8700

 

[topsquark]

$\quad\displaystyle \frac{dy}{dt}=2y-5, \quad y(0)=y_0$
rewrite
$\quad y'-2y=-5$
obtain u(x)
$\quad u(x)=\exp\int-2\, dx = e^{-2t}$
then
$\quad (e^{-2t}y)'=-5e^{-2t}$
integrate both sides
$\quad\displaystyle e^{-2t}y=-5\int e^{-2t} dt=-5e^{-2t}+c_1$
finally
$\quad\displaystyle y
=-5\frac{e^{-2t}}{e^{-2t}}+\frac{c_1}{e^{-2t}}=c_1e^{2t}-5$
so at $\quad y(0)=y_0$

ok I don't see how they got this answer

First, what is your c1 in terms of \(\displaystyle y_0\)? You never finished that part.

Also take a look at this:
\(\displaystyle y = 5 +(y_0 - 5)e^{-t}\)

\(\displaystyle y' = -(y_0 - 5)e^{-t}\)

Thus
\(\displaystyle y' - 2y = -(y_0 - 5)e^{-t} - 2(5 + (y_0 - 5)e^{-t}) = -10 - 3(y_0 - 5)e^{-t} \neq -5\)

So the given solution does not match the differential equation. You have the skill to do this check. (And you should always check your own solutions anyway.)

-Dan

 

[MarkFL]

We have:

\(\displaystyle \frac{d}{dt}\left(e^{-2t}y\right)=-5e^{-2t}\)

Integrate using the boundaries:

\(\displaystyle \int_{y_0}^{e^{-2t}y}\,du=-5\int_{0}^{t}e^{-2v}\,dv\)

\(\displaystyle e^{-2t}y-y_0=\frac{5}{2}\left(e^{-2t}-1\right)\)

\(\displaystyle y-y_0e^{2t}=\frac{5}{2}\left(1-e^{2t}\right)\)

\(\displaystyle y=\left(y_0-\frac{5}{2}\right)e^{2t}+\frac{5}{2}\)

 

[karush]

$\quad\displaystyle \frac{dy}{dt}=2y-5, \quad y(0)=y_0$
rewrite
$\quad y'-2y=-5$
obtain u(x)
$\quad u(x)=\exp\int-2\, dx = e^{-2t}$
then
$\quad (e^{-2t}y)'=-5e^{-2t}$
integrate both sides
$\quad\displaystyle e^{-2t}y=-5\int e^{-2t} dt=-5e^{-2t}+c_1$
finally
$\quad\displaystyle y
=-5\frac{e^{-2t}}{e^{-2t}}+\frac{c_1}{e^{-2t}}=c_1e^{2t}-5$
so at $\quad y(0)=y_0$
$\quad\displaystyle y(0)=-5+c_1=y_0 $
then
$\quad\displaystyle c_1=y_0+5$

 

[karush]

We have:

\(\displaystyle \frac{d}{dt}\left(e^{-2t}y\right)=-5e^{-2t}\)

Integrate using the boundaries:

\(\displaystyle \int_{y_0}^{e^{-2t}y}\,du=-5\int_{0}^{t}e^{-2v}\,dv\)

\(\displaystyle e^{-2t}y-y_0=\frac{5}{2}\left(e^{-2t}-1\right)\)

\(\displaystyle y-y_0e^{2t}=\frac{5}{2}\left(1-e^{2t}\right)\)

\(\displaystyle y=\left(y_0-\frac{5}{2}\right)e^{2t}+\frac{5}{2}\)

well that is pretty cool ... better than the book process
but I still don't think the boundary thing has registered with me

 

[topsquark]

well that is pretty cool ... better than the book process
but I still don't think the boundary thing has registered with me

When solving differential equations you see a lot of arbitrary constants. (ie. there are many solutions to the same differential equation.) The boundary conditions simply give you information about those constants. In this case you had c1 as a constant after solving the differential equation. The boundary condition \(\displaystyle y(0) = y_0\) tells you what c1 is in terms of \(\displaystyle y_0\). Perhaps it would be more meaningful to you if we specified y(0) = 3 or something like that.

-Dan