# DE 2.1.1.16 Find the solution of the give initial value problem

• MHB
• karush
In summary, the solution to the given initial value problem is $y(\pi)=0$, where $y(\pi)=\dfrac{\sin{\pi}}{\pi^2}+\dfrac{c}{\pi^2}=0$ if $y(\pi)eq 0$
karush
Gold Member
MHB
Find the solution of the give initial value problem

$\displaystyle y^\prime - \frac{2}{t}y =\frac{\cos{t}}{t^2}; \quad y{(\pi)}=0, \quad t>0$$u(t)=e^{2 \ln{t}}$then
$\displaystyle e^{2\ln{t}}\, y^\prime - \frac{2e^{e^{2\ln{t}}}}{t}y = \frac{e^{2\ln{t}}\cos{t}}{t^2}$not sure actually!

$e^{\int -\frac{2}{t} \, dt} = e^{-2\ln{t}} = \dfrac{1}{t^2}$

check the original DE statement again ...

skeeter said:
$e^{\int -\frac{2}{t} \, dt} = e^{-2\ln{t}} = \dfrac{1}{t^2}$

check the original DE statement again ...

thus...

$\displaystyle e^{-2\ln{t}}\, y^\prime - \frac{-2e^{e^{2\ln{t}}}}{t}y = \frac{e^{-2\ln{t}}\cos{t}}{t^2}$

that’s not what I meant ...

please double check the original DE from wherever you got it from, specifically the right side

skeeter said:
that’s not what I meant ...

please double check the original DE from wherever you got it from, specifically the right side
#16View attachment 9702

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karush said:
#16

First of all, please note you made a typo in your original DE, which is why Skeeter said your integrating factor is wrong.

With the correct equation as in your picture, your integrating factor is correct, but you will not be able to do the integration unless you simplify it. $\displaystyle \mathrm{e}^{2\ln{(t)}} = \mathrm{e}^{\ln{\left( t^2 \right) }} = t^2$. Then the integration will be doable...

Find the solution of the give initial value problem

$\displaystyle y^\prime + \frac{2}{t}y =\frac{\cos{t}}{t^2}; \quad y{(\pi)}=0, \quad t\ge 0$

$u(t)=e^{\displaystyle\ln{t^2}}$

then
$\displaystyle e^{\ln{t^2}}\, y^\prime +\frac{e^{\ln{t^2}}}{t}y = \frac{e^{\ln{t^2}}\cos{t}}{t^2}$

$\displaystyle(y'\cdot e^{\ln{t^2}})'=\frac{e^{\ln{t^2}}\cos{t}}{t^2}=\cos{ t}$

proceed ?

Last edited:
$y’ + \dfrac{2}{t} \cdot y = \dfrac{\cos{t}}{t^2}$

multiply evert term by the integrating factor, $e^{\int 2/t \, dt} = e^{2\ln{t}} = t^2$ ...

$y’t^2 + 2ty = \cos{t}$

$(yt^2)’ = \cos{t}$

now proceed

skeeter said:
$y’ + \dfrac{2}{t} \cdot y = \dfrac{\cos{t}}{t^2}$

multiply evert term by the integrating factor, $e^{\int 2/t \, dt} = e^{2\ln{t}} = t^2$ ...

$y’t^2 + 2ty = \cos{t}$

$(yt^2)’ = \cos{t}$

now proceed
$y(t)=\dfrac{\sin{t}}{t^2}+\frac{c}{t^2}$
So if $y(\pi)=0$ then
$y(\pi)=\dfrac{\sin{\pi}}{\pi^2}+\dfrac{c}{\pi^2}=0$
$c=-1$

Really??

Last edited:
$\displaystyle yt^2 = \int \cos{t} \, dt$

$yt^2 = \sin{t} + C$

$y = \dfrac{\sin{t}}{t^2} + \dfrac{C}{t^2}$

$y(\pi) = 0 \implies 0 = \dfrac{\sin(\pi)}{\pi^2} + \dfrac{C}{\pi^2} \implies C = 0$

$y = \dfrac{\sin{t}}{t^2}$

Ok saw my error

## 1. What is DE 2.1.1.16?

DE 2.1.1.16 refers to a specific type of differential equation, specifically a first-order linear ordinary differential equation. It is commonly used in mathematical modeling and has a variety of applications in fields such as physics, engineering, and economics.

## 2. What does it mean to "find the solution" of a DE 2.1.1.16 problem?

Finding the solution of a DE 2.1.1.16 problem means determining the function that satisfies the given differential equation and initial conditions. This function represents the relationship between the dependent and independent variables in the problem and can be used to make predictions or analyze the system.

## 3. What is an initial value problem?

An initial value problem is a type of differential equation that involves finding the solution that satisfies both the differential equation and a set of initial conditions. These initial conditions specify the value of the dependent variable at a particular point or set of points, allowing for a unique solution to be determined.

## 4. How is the solution of a DE 2.1.1.16 problem typically found?

The solution of a DE 2.1.1.16 problem is typically found using analytical methods, such as separation of variables or integrating factors. These methods involve manipulating the differential equation to isolate the dependent and independent variables and then integrating to find the solution function.

## 5. What are some real-world applications of DE 2.1.1.16 problems?

DE 2.1.1.16 problems have a wide range of applications in various fields, such as modeling population growth, analyzing electrical circuits, and predicting the behavior of chemical reactions. They are also commonly used in physics to describe the motion of objects under the influence of forces, such as in the equations of motion for a falling object.

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