# -1.1.1c Solve for the following initial value

• MHB
• karush
In summary, the conversation discusses rewriting an equation, obtaining the integrating factor, and using the boundaries to find the solution. The final solution is given as $y=5+(y_0-5)e^{-2t}$, with a possible sign error in the book's answer. Another method, the Method of Characteristics, is also mentioned as a higher level approach.
karush
Gold Member
MHB
$1.1.2c\\ \displaystyle \frac{dy}{dt}=2y-10, \qquad y(0)=y_0$
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

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

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

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

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

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

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

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

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

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

book answer should have a positive exponent ...

skeeter said:
$\dfrac{dy}{dt} = 2(y-5)$

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

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

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

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

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

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

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

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

book answer should have a positive exponent ...

oh this one?
View attachment 8707

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

$1.1.2c\\ \displaystyle \frac{dy}{dt}=2y-10, \qquad y(0)=y_0$
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 ?

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 $e^{2t}$ ...

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

karush said:
$1.1.2c\\ \displaystyle \frac{dy}{dt}=2y-10, \qquad y(0)=y_0$
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

## 1. What does "initial value" refer to in this context?

In this context, "initial value" refers to the starting point or condition of a mathematical equation or problem. It is the value for a given variable at the beginning of the problem.

## 2. How is this type of equation typically solved?

This type of equation, known as a first-order differential equation, is typically solved using techniques such as separation of variables, integrating factors, or the method of undetermined coefficients.

## 3. What is the significance of the "-1.1.1c" in this equation?

The "-1.1.1c" represents the specific equation or problem that is being solved. It is a unique identifier that helps to differentiate this problem from others and allows for easy reference.

## 4. What does it mean to "solve" for a value?

To "solve" for a value means to find the specific value or values of a variable that make the given equation or problem true. It is the process of finding the solution or solutions to a mathematical problem.

## 5. Can this initial value problem have multiple solutions?

Yes, it is possible for this initial value problem to have multiple solutions. This can occur when the equation or problem has multiple variables or when there are different starting points or conditions that can lead to different solutions.

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