# -b.1.2.2c initial value problem

• MHB
• karush
In summary, the student is saying that the solution y(0) = y_0 has a constant c1 associated with it, and that the boundary conditions give information about this c1.
karush
Gold Member
MHB
$\displaystyle \frac{dy}{dt}=2y-5, \quad y(0)=y_0$
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:
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:

$$\displaystyle -5e^{-2t}$$

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

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

We have:

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

Integrate using the boundaries:

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

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

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

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

$\quad\displaystyle \frac{dy}{dt}=2y-5, \quad y(0)=y_0$
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
$\quad\displaystyle c_1=y_0+5$

MarkFL said:
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

karush said:
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

## What is an initial value problem?

An initial value problem is a type of mathematical problem that involves finding a function or set of functions that satisfy a given set of conditions, typically at a specific point or set of points. These conditions typically include an initial value, such as a starting point or boundary condition, and a set of differential equations or other equations that describe the behavior of the function.

## What does the "-b.1.2.2c" notation refer to in the "-b.1.2.2c initial value problem"?

The "-b.1.2.2c" notation refers to a specific type of initial value problem that is used in mathematics and physics. This notation typically indicates that the problem involves a second-order ordinary differential equation with constant coefficients.

## What is the purpose of solving an initial value problem?

The purpose of solving an initial value problem is to determine the behavior of a function or set of functions at specific points or over a specific range of values. This can be useful in modeling physical systems, predicting future behavior, and understanding the underlying principles and relationships that govern a system.

## What are some common methods for solving "-b.1.2.2c initial value problems"?

Some common methods for solving "-b.1.2.2c initial value problems" include the Laplace transform method, the method of undetermined coefficients, and the variation of parameters method. These methods involve different techniques for manipulating and solving differential equations to find the desired functions.

## What are some applications of "-b.1.2.2c initial value problems" in the real world?

"-b.1.2.2c initial value problems" have many applications in the real world, including in physics, engineering, and economics. They can be used to model physical systems, such as the motion of objects or the flow of fluids, and to predict behavior in these systems. They are also commonly used in economic models to predict future trends and make informed decisions.

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