# Thread: Differential Equations HW question

1. ## Differential Equations HW question

Hey guys, been looking for some online support for diff eqs and this is where I wound up :P Just started summer session last week so I have a semester's worth of diff eqs crammed into 7 weeks and the HW is coming hard and heavy. This question I have is pretty elementary stuff but I am not very well-versed in diff eqs so I am having some trouble with it.

Here it is verbatim from the book:

a.) Draw a direction field for the given differential equation. How do solutions appear to behave as t becomes large? Does the behavior depend on the choice of the initial value a? Let a0 be the value of a for which the transition from one type of behavior to another occurs. Estimate the value of a0.

b.) Solve the initial value problem and find the critical value a0 exactly.

c.) Describe the behavior of the solution corresponding to the initial value a0.

Equation: y' - y/2 = 2cost, y(0) = a

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Now, I can draw the direction field, albeit it's a little time-consuming. I can solve the equation as well since it's linear; not too tough. The problems I have are with the initial value stuff. I just plain don't understand how to do it.

Does the behavior depend on the initial value a? I don't know, I don't think so. I'm not sure how to tell. Estimate the value of a0? I have no idea how to do this either, other than looking at my direction field and just guessing. Find the critical value a0? Don't know how to do that one.

I mean, when you solve the equation you're going to be left with a constand C and then you have your initial value y(0) = a = (some equation + C). How are you supposed to solve for a when you have an unknown C?

Anyway, this stuff probably isn't hard but it's more or less my first experience with it, so I just need to get my mind wrapped around it a little bit I think. Hopefully someone around here can help me out. Thanks guys.

2. Originally Posted by griffsterb
Hey guys, been looking for some online support for diff eqs and this is where I wound up :P Just started summer session last week so I have a semester's worth of diff eqs crammed into 7 weeks and the HW is coming hard and heavy. This question I have is pretty elementary stuff but I am not very well-versed in diff eqs so I am having some trouble with it.

Here it is verbatim from the book:

a.) Draw a direction field for the given differential equation. How do solutions appear to behave as t becomes large? Does the behavior depend on the choice of the initial value a? Let a0 be the value of a for which the transition from one type of behavior to another occurs. Estimate the value of a0.

b.) Solve the initial value problem and find the critical value a0 exactly.

c.) Describe the behavior of the solution corresponding to the initial value a0.

Equation: y' - y/2 = 2cost, y(0) = a

----------

Now, I can draw the direction field, albeit it's a little time-consuming. I can solve the equation as well since it's linear; not too tough. The problems I have are with the initial value stuff. I just plain don't understand how to do it.

Does the behavior depend on the initial value a? I don't know, I don't think so. I'm not sure how to tell. Estimate the value of a0? I have no idea how to do this either, other than looking at my direction field and just guessing. Find the critical value a0? Don't know how to do that one.

I mean, when you solve the equation you're going to be left with a constand C and then you have your initial value y(0) = a = (some equation + C). How are you supposed to solve for a when you have an unknown C?

Anyway, this stuff probably isn't hard but it's more or less my first experience with it, so I just need to get my mind wrapped around it a little bit I think. Hopefully someone around here can help me out. Thanks guys.

So you know that the general solution to the DE is:

$\displaystyle y(t)=A \exp(t/2)-(4/5)\cos(t)+(8/5)\sin(t)$

Also you have the initial condition $\displaystyle y(0)=a$, but:

$\displaystyle y(0)=A -(4/5)$

So $\displaystyle A=a+(4/5)$, and the solution is:

$\displaystyle y(t)=[a+(4/5)] \exp(t/2)-(4/5)\cos(t)+(8/5)\sin(t)$

RonL

3. Okay thanks for your help. I can see how it actually works out. However, working through the problem I got stuck on the integral when solving the equation.

The method we're using for linear equations right now is that $\displaystyle y(t) = [1/mu(y)] * integral[mu(y)*g(t)]dt$

With linear equations of the form $\displaystyle y' + mu(y) = g(t)$

So eventually it comes to $\displaystyle y(t) = exp(t/2) * integral[2cos(t)/exp(t/2)]dt$

And I'm not sure how to solve that integral in there. Integration by parts won't work and neither will any of the more simpler methods that I recall from Calc. It's been a while though since I did any complicated integrations so I'm probably forgetting something =\

And also: Is there a way to insert math symbols into my posts? The integral symbol makes things look cleaner than writing "integral" :P

4. Originally Posted by griffsterb
Okay thanks for your help. I can see how it actually works out. However, working through the problem I got stuck on the integral when solving the equation.

The method we're using for linear equations right now is that $\displaystyle y(t) = [1/mu(y)] * integral[mu(y)*g(t)]dt$

With linear equations of the form $\displaystyle y' + mu(y) = g(t)$

So eventually it comes to $\displaystyle y(t) = exp(t/2) * integral[2cos(t)/exp(t/2)]dt$

And I'm not sure how to solve that integral in there. Integration by parts won't work and neither will any of the more simpler methods that I recall from Calc. It's been a while though since I did any complicated integrations so I'm probably forgetting something =\

And also: Is there a way to insert math symbols into my posts? The integral symbol makes things look cleaner than writing "integral" :P

I think this is the integral you want

$\displaystyle \int \frac{2\cos(t)}{e^{\frac{t}{2}}}dt$

Here is the la Tex for it \int \frac{2\cos(t)}{e^{\frac{t}{2}}}dt

Here is a link with all the code you will need

Helpisplaying a formula - Wikipedia, the free encyclopedia

If so integration by parts will work it just needs another little "trick"

First lets rewrite this as

$\displaystyle 2\int e^{\frac{-t}{2}}\cos(t)dt$

by parts with $\displaystyle u=e^{\frac{-t}{2}} \\\ dv=\cos(t)$

we get

$\displaystyle 2\int e^{\frac{-t}{2}}\cos(t)dt=e^{\frac{-t}{2}}\sin(t)+\frac{1}{2}\int e^{\frac{-t}{2}}\sin(t)dt$

Now by parts again (I know you think this is going nowhere)

$\displaystyle u=e^{\frac{-t}{2}} \\\ dv=\sin(t)$ we get

$\displaystyle 2\int e^{\frac{-t}{2}}\cos(t)dt=e^{\frac{-t}{2}}\sin(t)+\frac{1}{2}\left( -e^{\frac{-t}{2}}\cos(t)-\frac{1}{2}\int e^{\frac{-t}{2}}\cos(t)dt\right)$

Simplifying we get

$\displaystyle 2\int e^{\frac{-t}{2}}\cos(t)dt=e^{\frac{-t}{2}}\sin(t)-\frac{1}{2}e^{\frac{-t}{2}}\cos(t)-\frac{1}{4}\int e^{\frac{-t}{2}}\cos(t)dt$

Now this is where the "magic" happens notice we have the same integral on both sides of the equation so we add $\displaystyle \frac{1}{4}\int e^{\frac{-t}{2}}\cos(t)dt$ to both sides of the equation to get

$\displaystyle \frac{9}{4}\int e^{\frac{-t}{2}}\cos(t)dt=e^{\frac{-t}{2}}\sin(t)-\frac{1}{2}e^{\frac{-t}{2}}\cos(t)$

Multiplying by 4/9 and factoring gives

$\displaystyle \int e^{\frac{-t}{2}}\cos(t)dt=\frac{e^{\frac{-t}{2}}\left(4\sin(t)-2\cos(t)\right)}{9}$

P.S. Most people just look this up in an integral table.

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### differential equations what happens when t becomes large

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