- #1
Xyius
- 508
- 4
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..
[tex]y(t)=c_{1}cos(t)+c_{2}sin(t)+\int^{t}_{0}sin(t-s)ds[/tex]
.. is the general solution to the equation..
[tex]y''+y'=f(t)[/tex]
Where f(t) is a continuous function on the real number line. (HINT: Use the trigonometric identity [tex]sin(t-s)=sin(t)cos(s)-sin(s)cos(t)[/tex].
The way I solved it was to first find the homogeneous solution and then use variation of parameters to obtain..
[tex]y_{p}(t)=v_{1}cos(t)+v_{2}sin(t)[/tex]
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!
The question says:
Use the method of variation of parameters to show that..
[tex]y(t)=c_{1}cos(t)+c_{2}sin(t)+\int^{t}_{0}sin(t-s)ds[/tex]
.. is the general solution to the equation..
[tex]y''+y'=f(t)[/tex]
Where f(t) is a continuous function on the real number line. (HINT: Use the trigonometric identity [tex]sin(t-s)=sin(t)cos(s)-sin(s)cos(t)[/tex].
The way I solved it was to first find the homogeneous solution and then use variation of parameters to obtain..
[tex]y_{p}(t)=v_{1}cos(t)+v_{2}sin(t)[/tex]
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!