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[SOLVED] b.1.1.2 initial value problem

karush

Well-known member
Jan 31, 2012
2,886
$$\displaystyle \frac{dy}{dt}=2y-5, \quad y(0)=y_0$$
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
?transparentcup.png
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
Jan 31, 2012
2,886
$\quad\displaystyle \frac{dy}{dt}=2y-5, \quad y(0)=y_0$
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

2.1.PNG
 
Last edited:

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,180
$\quad\displaystyle \frac{dy}{dt}=2y-5, \quad y(0)=y_0$
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

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
Jan 31, 2012
2,886
$\quad\displaystyle \frac{dy}{dt}=2y-5, \quad y(0)=y_0$
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

Well-known member
Jan 31, 2012
2,886
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-known member
MHB Math Helper
Aug 30, 2012
1,180
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