How do you solve this simple differential equation?

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Homework Help Overview

The discussion revolves around solving a differential equation involving an integral of a function v(t). The equation presented is 2 = v(t)/10 + (1/2)∫v(t) dt, which raises questions about the methods for solving such equations and the implications of constants in the solution.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants suggest rearranging the equation and differentiating to eliminate the integral. There is mention of a specific form for the solution involving exponential functions, and questions arise about finding constants without initial conditions.

Discussion Status

The discussion is active with participants exploring different approaches to the problem. Some guidance has been offered regarding rearranging the equation and the form of the solution, but there is no explicit consensus on the method or final outcome.

Contextual Notes

Participants note the challenge of solving the equation due to the presence of a constant rather than a function, and there is uncertainty about the implications of initial conditions on finding constants in the solution.

elcotufa
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Homework Statement



How would you solve

2=\frac{v(t)}{10}+\frac{\int{v(t)}}2


Thanks for your help
 
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elcotufa said:

Homework Statement



How would you solve

2=\frac{v(t)}{10}+\frac{\int{v(t)}}2


Thanks for your help

Try rearranging the equation.

\int{v_{(t)} = 4 - \frac{v_{(t)}}{5}

What function do you know that will yield this kind of result?
 
I know the answer is 20e^(-5t) by just taking the 1/10 out from the v(t), and the k on top of the exponential is negative 10/2 but I don't know the solving mechanismMy calc book only has one example and it is when it equals a function and not a constant in the first equation, so I can just differentiate the equation to get rid of the integral

The answer should be in the form A+Bexp(-kt)
A is zero for this equation but can I find B without any initial conditions?
 
elcotufa said:
I know the answer is 20e^(-5t) by just taking the 1/10 out from the v(t), and the k on top of the exponential is negative 10/2 but I don't know the solving mechanism


My calc book only has one example and it is when it equals a function and not a constant in the first equation, so I can just differentiate the equation to get rid of the integral

The answer should be in the form A+Bexp(-kt)
A is zero for this equation but can I find B without any initial conditions?

\int{v_{(t)} = 4 - \frac{v_{(t)}}{5}

Taking the common integral of e you can start with this

\int{e^{(ct)} = \frac{e^{ct}}{c}

http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions
 

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