Initial value problem question

[yecko]

Homework Statement

Homework Equations

$y(t)\mu(t) - y(t_0) \mu(t_0) = \int_{t_0}^t \mu(s) g(s) ds$
$y(t) = \frac{1}{\mu(t)} \left[y_0 \mu(t_0) + \int_{t_0}^t \mu(s)g(s) ds\right]$

The Attempt at a Solution

(7 lines)I have done the first part, which seems correct, yet I am stuck with the second part although I have attempt so many times. I have got no idea with what is wrong with my attempt. Can anyone help?
Thanks for any help!

 

[Gene Naden]

Thanks for posting your question. In the "relevant equations", what is $\mu (t)$?

 

[yecko]

Integration factor
Calculated in the photo

 

[Ray Vickson]

Integration factor
Calculated in the photo

Many helpers will not look at the photo. Take the trouble to type it out here if you are really serious about wanting help.

 

[Mark44]

I have done the first part, which seems correct, yet I am stuck with the second part although I have attempt so many times.

Did you check your 2nd solution? In other words, did you verify that it satisfies the diff. equation? I did, and it looks like you have a sign wrong somewhere.

Also, you should simplify things a bit. $e^{-\cos(t) -1} = e^{-\cos(t)}\cdot e^{-1} = e^{-1}e^{-\cos(t)}$

 

[Gene Naden]

I have solved this problem using the formulas in Wikipedia, at https://en.wikipedia.org/wiki/Integrating_factor.

My answer does not agree with what you have posted for the instructors solution, nor does it agree with your solution. My solution for x>$\pi$ is:
$y(t) = (2 + 7e^{-2}) e^{(cos(t)+1)}-1$
Could you show more steps in your solution? Thanks

 

[yecko]

I have solved this problem using the formulas in Wikipedia, at https://en.wikipedia.org/wiki/Integrating_factor.

My answer does not agree with what you have posted for the instructors solution, nor does it agree with your solution. My solution for x>$\pi$ is:
$y(t) = (2 + 7e^{-2}) e^{(cos(t)+1)}-1$
Could you show more steps in your solution? Thanks

Your solution (i.e. (2+7e^(-2))e^(cos(t)+1)-1) should be correct...
Do you mind to share how can you obtain the answer? because I still have no idea on starting with which step am I wrong...
Thank you.

Many helpers will not look at the photo. Take the trouble to type it out here if you are really serious about wanting help.

My attempt (incorrect):
μ(t)=e^[∫ {from pi to t} -sin(t) dt]=e^(cost+1)
y(pi)=e^(-2)*(e^2+7)=1+7e^-2
μ(pi)=e^0=1
y(t)=e^(-cost-1)*[1+7e^-2+∫ {from pi to t} e^(cost+1)(-sint)dt] = e^(-cost-1)[7e^-2 + e^(cost+1)]

Did you check your 2nd solution? In other words, did you verify that it satisfies the diff. equation? I did, and it looks like you have a sign wrong somewhere.

Also, you should simplify things a bit. $e^{-\cos(t) -1} = e^{-\cos(t)}\cdot e^{-1} = e^{-1}e^{-\cos(t)}$

My answer seems nothing similar to the correct answer, and I have double checked the signs and I feel like they are fine... can anyone help? or to show your attempt for comparison?
Thanks

 

[Gene Naden]

Hello Yecko,
As I said, I used the formulas form Wikipedia. There is a link to the page on my earlier post.
The differential equation is $y\prime + Py = Q$ where P=sin(t) and, for t>$\pi$, Q=-sin(t).
The solution is $y(t)=A(t)B(t)+CD(t)$ where C is a constant of integration, to be determined by the boundary condition $\pi = 1+7e^{-2}$.
$A(x)=e^{-\int_{\pi}^x P(s) ds}$
$B(x)=\int_{\pi}^x Q(t)e^{\int_\pi ^t P(s)ds}dt$
and $D(x)=e^-{\int _\pi ^x P(s) ds}$
I suggest you try to do these integrals. Maybe you can relate them to your formula $\mu(t)=e^{\int _\pi ^t Q(t) ds}$
Let me know how far you get... thanks.

 

[Gene Naden]

Oh, yes, to do one of the integrals I used the online integrator at http://www.wolframalpha.com/

 

[ehild]

Using definite integrals over-complicates the problem.
The integrating factor is I=e^-cos(t), and $Iy=\int{Igdt}$ Include integration constant, and match it to the initial conditions for both parts of y(t).

 

[Gene Naden]

I would very much like to see a little more about how to use the integrating factor. I agree that my solution is very complicated and would welcome an opportunity to simplify it!

 

[ehild]

I have solved this problem using the formulas in Wikipedia, at https://en.wikipedia.org/wiki/Integrating_factor.

My answer does not agree with what you have posted for the instructors solution, nor does it agree with your solution. My solution for x>$\pi$ is:
$y(t) = (2 + 7e^{-2}) e^{(cos(t)+1)}-1$
Could you show more steps in your solution? Thanks

Your solution is correct. Check this : http://www.mathcentre.ac.uk/resources/uploaded/mathcentre-ode.pdf

 

[Gene Naden]

What I meant to say was that my method, doing all the definite integrals, was very complicated and I would welcome knowing about a simpler method.

 

[ehild]

See the URL in my previous Post. You do not need definite integrals.
There is an other method which is simple for this problem.
You have linear first-order differential equations for the two domains.The homogeneous parts (dy/dt+sin(t)y)) are the same, the inhomogeneous parts differ.(sin(t), -sin(t).
Do you know that the solution of a linear equation is Y=Yh + Yp, where Yh is the general solution of the homogeneous equation (dy/dt+sin(t)y)=0) and Yp is a particular solution of the inhomogeneous equation (dy/dt+sin(t)y)=g(t)). Yp happens to be very simple, a constant function for both parts.

 

[Ray Vickson]

What I meant to say was that my method, doing all the definite integrals, was very complicated and I would welcome knowing about a simpler method.

Is your question asking how to choose the integrating factor? In general, the solution of $$dy/dx + f(x) y = g(x)$$ on $x \geq 0$ has the form
$$y(x) = c e^{-F(x)} + e^{-F(x)} \int_0^x e^{F(y)} g(y) \, dy,$$
where
$$F(x) = \int_a^x f(t) \, dt$$
If we change the lower limit '$a$' we change $F$ as well; however, that does not affect the solution $y(x)!$ To see this, just change $F$ to $F+k$ for some constant $k$. The new $y$ is
$$y_{\text{new}}(x) = c \,e^{-k} e^{-F(x)} + e^{-k} e^{-F(x)} \int_0^x e^{k} e^{F(y)} g(y) \, dy$$
The factors $e^{-k}$ and $e^k$ cancel in the integral term, and the non-integral term just has a different constant $c_1 = c \,e^{-k}$. Since you determine the constant to match the initial condition (or whatever type of condition you are given), you get the same solution in the end.

So: choose any antiderivatve $\int fx) \, dx$ and forget about the constant of integration. The only thing that changes is the coefficient of the constant term (the $c$ in our formulas above).

 

[Gene Naden]

This is very helpful. I particularly liked the splitting of $y$ into $y_h + y_p$. When I did that, I didn't have to do any integrals at all except to solve
$\frac{d y_h}{dt}+y_h sin(t)=0$!