2.1.1c Solve for the following initial value

karush

Well-known member
$1.1.2c\\ \displaystyle \frac{dy}{dt}=2y-10, \qquad y(0)=y_0$
rewrite
$\quad\displaystyle y'-2y=-10$
obtain u(x)
$\quad\displaystyle u(t)=\exp\int -2 \, dt=e^{-2t}$
integrate using the boundaries:
$\quad\displaystyle \int_{y_0}^{e^{-2t}y}\,du=-10\int_{0}^{t}e^{-2v}\,dv$
resulting in
$\quad\displaystyle e^{-2t}y-y_0=-5\left(e^{-2t}-1\right)$
multiply thru by $e^{2t}$
$\quad\displaystyle y-y_0e^{2t}=5\left(1-e^{2t}\right)$
then
$\quad\displaystyle y=5\left(1-e^{2t}\right)+y_0e^{2t}$
then
$\quad\displaystyle y=5-5e^{2t}+y_0e^{2t}$
then
$\quad\displaystyle y=5-(y_0+5)e^{-2t}$
$\quad\displaystyle y=5+(y_0-5)e^{-2t}$

ok seems to be a sign error someplace

skeeter

Well-known member
MHB Math Helper
$\dfrac{dy}{dt} = 2(y-5)$

$\dfrac{dy}{y-5} = 2 \,dt$

$\ln|y-5| = 2t+ C$

$y(0) = y_0 \implies \ln|y_0-5| = C$

$\ln|y-5| = 2t+ \ln|y_0-5|$

$\ln \bigg| \dfrac{y-5}{y_0-5} \bigg| = 2t$

$\dfrac{y-5}{y_0-5} = e^{2t}$

$y-5 = (y_0-5)e^{2t}$

$y = 5 + (y_0-5)e^{2t}$

book answer should have a positive exponent ...

karush

Well-known member
$\dfrac{dy}{dt} = 2(y-5)$

$\dfrac{dy}{y-5} = 2 \,dt$

$\ln|y-5| = 2t+ C$

$y(0) = y_0 \implies \ln|y_0-5| = C$

$\ln|y-5| = 2t+ \ln|y_0-5|$

$\ln \bigg| \dfrac{y-5}{y_0-5} \bigg| = 2t$

$\dfrac{y-5}{y_0-5} = e^{2t}$

$y-5 = (y_0-5)e^{2t}$

$y = 5 + (y_0-5)e^{2t}$

book answer should have a positive exponent ...
oh this one?
View attachment 8707

so you did this without the $u(x)$ method!

karush

Well-known member
$1.1.2c\\ \displaystyle \frac{dy}{dt}=2y-10, \qquad y(0)=y_0$
rewrite
$\quad\displaystyle y'-2y=-10$
obtain u(x)
$\quad\displaystyle u(t)=\exp\int -2 \, dt=e^{-2t}$
integrate using the boundaries:
$\quad\displaystyle \int_{y_0}^{e^{-2t}y}\,du=-10\int_{0}^{t}e^{-2v}\,dv$
resulting in
$\quad\displaystyle e^{-2t}y-y_0=-5\left(e^{-2t}-1\right)$
multiply thru by $e^{2t}$
$\quad\displaystyle y-y_0e^{2t}=-5\left(1-e^{2t}\right)$
then
$\quad\displaystyle y=-5\left(1-e^{2t}\right)+y_0e^{2t}$
then
$\quad\displaystyle y=-5+5e^{2t}+y_0e^{2t}$
then
$\quad\displaystyle y=5-(y_0+5)e^{2t}$
$\quad\displaystyle y=5+(y_0-5)e^{-2t}$

still sign ???

skeeter

Well-known member
MHB Math Helper
using an integrating factor ...

$\dfrac{dy}{dt} = 2y-10$

$\dfrac{dy}{dt} - 2y = -10$

$\mu(t) = e^{-2t}$

$e^{-2t}\left(\dfrac{dy}{dt} - 2y\right) = -10e^{-2t}$

$\dfrac{d}{dt}\left(e^{-2t} \cdot y\right) = -10e^{-2t}$

integrate both sides ...

$e^{-2t} \cdot y = 5e^{-2t} + C$

$y(0) = y_0 \implies y_0 = 5 + C \implies C = y_0 - 5$

$e^{-2t} \cdot y = 5e^{-2t} + y_0 - 5$

multiply every term by $e^{2t}$ ...

$y = 5 + (y_0 - 5)e^{2t}$

topsquark

Well-known member
MHB Math Helper
$1.1.2c\\ \displaystyle \frac{dy}{dt}=2y-10, \qquad y(0)=y_0$
rewrite
$\quad\displaystyle y'-2y=-10$
obtain u(x)
$\quad\displaystyle u(t)=\exp\int -2 \, dt=e^{-2t}$
integrate using the boundaries:
$\quad\displaystyle \int_{y_0}^{e^{-2t}y}\,du=-10\int_{0}^{t}e^{-2v}\,dv$
resulting in
$\quad\displaystyle e^{-2t}y-y_0=-5\left(e^{-2t}-1\right)$
multiply thru by $e^{2t}$
$\quad\displaystyle y-y_0e^{2t}=-5\left(1-e^{2t}\right)$
then
$\quad\displaystyle y=-5\left(1-e^{2t}\right)+y_0e^{2t}$
then
$\quad\displaystyle y=-5+5e^{2t}+y_0e^{2t}$
then
$\quad\displaystyle y=5-(y_0+5)e^{2t}$
$\quad\displaystyle y=5+(y_0-5)e^{-2t}$

still sign ???
In my book $$\displaystyle \dfrac{-10}{-2} = 5$$.

Here's a higher brow method, used when we have a linear nth order equation. (It's kinda overkill in this situation.)

Method of Characteristics:
$$\displaystyle y' - 2y = -10$$

Homogeneous equation:
$$\displaystyle y' - 2y = 0$$

Thus $$\displaystyle m - 2 = 0 \implies m = 2$$

So the homogeneous solution is $$\displaystyle y = Ae^{2t}$$

For the particular solution we take y = B
$$\displaystyle y' = 0$$

So
$$\displaystyle y' - 2y = -10 \implies 0 - 2(B) = -10 \implies B = 5$$

Thus for the whole equation
$$\displaystyle y = A e^{2t} + 5$$

(Then use $$\displaystyle y(0) = y_0$$ to find A.)

-Dan