Solution to a DE using variation of parameters

[Xyius]

I was looking through my DE book and a problem intrigued me. I eventually figured it out but I do not understand the logic. I was wondering if anyone here could help me out.

The question says:
Use the method of variation of parameters to show that..

y(t)=c_{1}cos(t)+c_{2}sin(t)+\int^{t}_{0}sin(t-s)ds

.. is the general solution to the equation..

y''+y'=f(t)

Where f(t) is a continuous function on the real number line. (HINT: Use the trigonometric identity sin(t-s)=sin(t)cos(s)-sin(s)cos(t).

The way I solved it was to first find the homogeneous solution and then use variation of parameters to obtain..

y_{p}(t)=v_{1}cos(t)+v_{2}sin(t)

The odd part that I do not understand, is I had to make both "v" functions in terms of "s" to obtain the correct form of the solution.

The whole concept of using a definite integral in a differential equation is foreign to me and I do not understand it.

If anyone can help me understand this with a solid logic behind the solution it would be GREATLY appreciated. Thanks!

 

[LCKurtz]

I was looking through my DE book and a problem intrigued me. I eventually figured it out but I do not understand the logic. I was wondering if anyone here could help me out.

The question says:
Use the method of variation of parameters to show that..

y(t)=c_{1}cos(t)+c_{2}sin(t)+\int^{t}_{0}sin(t-s)ds

.. is the general solution to the equation..

Surely you have left the function f out of that answer. Probably an f(s) in the integrand.

y''+y'=f(t)

And also that is supposed to be y'' + y = f(t)

Where f(t) is a continuous function on the real number line. (HINT: Use the trigonometric identity sin(t-s)=sin(t)cos(s)-sin(s)cos(t).

The way I solved it was to first find the homogeneous solution and then use variation of parameters to obtain..

y_{p}(t)=v_{1}cos(t)+v_{2}sin(t)

The odd part that I do not understand, is I had to make both "v" functions in terms of "s" to obtain the correct form of the solution.

The whole concept of using a definite integral in a differential equation is foreign to me and I do not understand it.

If anyone can help me understand this with a solid logic behind the solution it would be GREATLY appreciated. Thanks!

You didn't show your work but I'm guessing you got these two equations:

v_1' = -f(t)\sin(t)

v_2' = f(t)\cos(t)

and you are confused by writing

v_1(t) = \int_0^t -f(s)\sin(s)\, ds

v_2(t) = \int_0^t f(s)\cos(s)\, ds

What you need to notice is that the variable s under the integral is a dummy variable. It doesn't matter what letter you use there because when you take the antiderivative you are going to substitute the upper and lower limits in for s anyway. These integrals are functions of t because of the upper limit. And by the fundamental theorem of calculus, if you differentiate either of them with respect to t, you just get the integrand with t substituted in for the s, so they are correct antiderivatives.

When you put them in your y_p you get the required formula.

 

[Xyius]

Wow I apologize for all the mistakes! It was 2 am! Ha!

Thank you very much you have helped a lot but I still do not understand the point of introducing the variable "s". If it is incorrect to write the "v" functions in terms of "s" then when I plug everything in I do not have any s's in the final solution.

Also I do not understand one other thing you said. Why will I obtain the integrand if I differentiate the integral with respect to t? The differential is ds. I DO understand what you said that the integral is actually a a function of t because of the limits. Somehow things aren't adding up for me. :\

 

[LCKurtz]

Wow I apologize for all the mistakes! It was 2 am! Ha!

Thank you very much you have helped a lot but I still do not understand the point of introducing the variable "s". If it is incorrect to write the "v" functions in terms of "s" then when I plug everything in I do not have any s's in the final solution.

Also I do not understand one other thing you said. Why will I obtain the integrand if I differentiate the integral with respect to t? The differential is ds. I DO understand what you said that the integral is actually a a function of t because of the limits. Somehow things aren't adding up for me. :\

It doesn't matter what letter you use for the integration variable. You could even use t but that is a bad idea because it causes confusion between it and the upper limit.

Look at this example. Say you have a function f(t) and you have taken its antiderivative and let's call it F(t), so F'(t) = f(t). Now, if you were doing a definite integral you would write

F(b) - F(a) = \int_a^b f(t)\, dt

And you get the same thing using s for the dummy variable:

F(b) - F(a) = \int_a^b f(s)\, ds

Now let's let b = t and a = 0 and we can write

F(t) = F(0) + \int_0^t f(s)\, ds

Now look what happens if you differentiate both sides with respect to t. We know F'(t) = f(t) from above. F(0) is a constant which goes away. So you have the derivative of the integral as a function of its upper limit on the right:

F'(t) = f(t) = \frac d {dt}\int_0^t f(s)\, ds

That's why you get f(t) when you differentiate the integral as a function of its upper limit.

 

[Xyius]

Ohh okay! Makes perfect sense!

Lemme just make sure I really nail this down. Does that mean I can write the integral

F(x) = \int f(x)cos(x)dx

as

\int ^{x}_{\frac{\pi}{2}}f(s)cos(s)ds

because

\int ^{x}_{\frac{\pi}{2}}f(s)cos(s)ds = F(x)-F(\frac{\pi}{2})

and

F(\frac{\pi}{2}) = \int f(\frac{\pi}{2})cos(\frac{\pi}{2})dx = 0

 

[LCKurtz]

Ohh okay! Makes perfect sense!

Lemme just make sure I really nail this down. Does that mean I can write the integral

F(x) = \int f(x)cos(x)dx

as

\int ^{x}_{\frac{\pi}{2}}f(s)cos(s)ds

because

\int ^{x}_{\frac{\pi}{2}}f(s)cos(s)ds = F(x)-F(\frac{\pi}{2})

and

F(\frac{\pi}{2}) = \int f(\frac{\pi}{2})cos(\frac{\pi}{2})dx = 0

No, its best not to mix up indefinite integrals with definite ones. When you write

F(t) - F(a) = \int_a^t f(s)\, ds

F(t) = \int_a^t f(s)\, ds + F(a)

the F(a) is like the constant of integration in an indefinite integral. If you change a to a different value c, the two expressions for F(t) will differ by a constant. In your original problem, you were looking for a particular solution so it didn't matter what the lower limit was. 0 was just a convenient choice.

 

[Xyius]

Cool! Thank you very much I understand completely now :D