# [SOLVED]b.1.1.2 initial value problem

#### karush

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

Last edited:

#### MarkFL

Staff member
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
$\quad\displaystyle \frac{dy}{dt}=2y-5, \quad y(0)=y_0$
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 Last edited:

#### topsquark

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

Staff member
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
$\quad\displaystyle \frac{dy}{dt}=2y-5, \quad y(0)=y_0$
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
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
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