Finding the Particular Integral for d2y/dt2 + 4y = 5sin2t

[VoteSaxon]

Homework Statement
Well I am looking for the particular integral of:
^d2y/_dt² + 4y = 5sin2t

The attempt at a solution
As f(t) = 5sin2t, the particular integral y_PI should look like:
y_PI = Acos2t + Bsin2t
^dy_PI/_dt = -2Asin2t + 2Bcos2t
^d2y_PI/_dt² = -4Acos2t - 4Bsin2t

Subbing into the differential equation, you get:
-4Acos2t - 4Bsin2t + 4(Acos2t + Bsin2t) = 5sin2t
cancelling ...
0 = 5sin2t

Surely this is incorrect. Any help please?

PS: I think I have already calculated the complementary function y_CF, so to get the overall solution I need the particular integral, as y_GEN = y_CF + y_PI

Edit: sorry for posting in the wrong forum, must have taken a wrong turn somewhere. Thanks for putting it right.

 

[SteamKing]

Homework Statement
Well I am looking for the particular integral of:
^d2y/_dt² + 4y = 5sin2t

The attempt at a solution
As f(t) = 5sin2t, the particular integral y_PI should look like:
y_PI = Acos2t + Bsin2t
^dy_PI/_dt = -2Asin2t + 2Bcos2t
^d2y_PI/_dt² = -4Acos2t - 4Bsin2t

Subbing into the differential equation, you get:
-4Acos2t - 4Bsin2t + 4(Acos2t + Bsin2t) = 5sin2t
cancelling ...
0 = 5sin2t

Surely this is incorrect. Any help please?

Since the homogeneous solution, B sin (2t), is similar to the RHS, 5 sin (2t), you should assume a particular solution of the form Bt sin (2t), as explained here:

http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

(See all the way at the bottom of the page.)

Also, you should post your HW in the appropriate forum. Differential equations fall under the Calculus HW forum.

 

[Mark44]

Please post questions about calculus and differential equations in the Calculus & Beyond section, not in the Precalc section. I have moved your thread.

Homework Statement
Well I am looking for the particular integral of:
^d2y/_dt² + 4y = 5sin2t

The attempt at a solution
As f(t) = 5sin2t, the particular integral y_PI should look like:
y_PI = Acos2t + Bsin2t

No. This is the general solution to the homogeneous diff. equation: y'' + 4y = 0. The characteristic equation (assuming you have come across this term in your studies to date) is r² + 4 = 0, with solutions $r = \pm 2i$. So a fundamental set of solutions to the homogeneous equation turns out to be $y = e^{\pm 2i}$. This can be manipulated via Euler's Theorem (I think that's the one) to give a different set; namely {cos(2t), sin(2t)}.

^dy_PI/_dt = -2Asin2t + 2Bcos2t
^d2y_PI/_dt² = -4Acos2t - 4Bsin2t

Subbing into the differential equation, you get:
-4Acos2t - 4Bsin2t + 4(Acos2t + Bsin2t) = 5sin2t
cancelling ...
0 = 5sin2t

Surely this is incorrect. Any help please?

Have you come across any discussion in your textbook about how to handle a forcing function (the function on the right side of your DE) that is the same as or a constant multiple of one of your complementary functions?

PS: I think I have already calculated the complementary function y_CF, so to get the overall solution I need the particular integral, as y_GEN = y_CF + y_PI

 

[VoteSaxon]

Please post questions about calculus and differential equations in the Calculus & Beyond section, not in the Precalc section. I have moved your thread.No. This is the general solution to the homogeneous diff. equation: y'' + 4y = 0. The characteristic equation (assuming you have come across this term in your studies to date) is r² + 4 = 0, with solutions $r = \pm 2i$. So a fundamental set of solutions to the homogeneous equation turns out to be $y = e^{\pm 2i}$. This can be manipulated via Euler's Theorem (I think that's the one) to give a different set; namely {cos(2t), sin(2t)}.
Have you come across any discussion in your textbook about how to handle a forcing function (the function on the right side of your DE) that is the same as or a constant multiple of one of your complementary functions?

Yes, for the complementary function I did get $y = e^{\pm 2i}$ Does this mean this is a sufficient solution to the equation? And about your last sentence, sorry, I do not think I have come across that in my studies.

 

[SteamKing]

Yes, for the complementary function I did get $y = e^{\pm 2i}$ Does this mean this is a sufficient solution to the equation? And about your last sentence, sorry, I do not think I have come across that in my studies.

Please see Post #2 above.

 

[Ray Vickson]

Homework Statement
Well I am looking for the particular integral of:
^d2y/_dt² + 4y = 5sin2t

The attempt at a solution
As f(t) = 5sin2t, the particular integral y_PI should look like:
y_PI = Acos2t + Bsin2t
^dy_PI/_dt = -2Asin2t + 2Bcos2t
^d2y_PI/_dt² = -4Acos2t - 4Bsin2t

Subbing into the differential equation, you get:
-4Acos2t - 4Bsin2t + 4(Acos2t + Bsin2t) = 5sin2t
cancelling ...
0 = 5sin2t

Surely this is incorrect. Any help please?

PS: I think I have already calculated the complementary function y_CF, so to get the overall solution I need the particular integral, as y_GEN = y_CF + y_PI

Edit: sorry for posting in the wrong forum, must have taken a wrong turn somewhere. Thanks for putting it right.

The particular solution should be $y_p(t) = A(t) \cos(2t) + B(t) \sin(2t)$. Substitute that into the DE and equate coefficients of $\sin(2t), \cos(2t)$ on both sides; that will give two coupled 2nd order differential equations for $A(t), B(t)$. These are actually coupled first-order DEs in the functions $A_1(t) = d\,A(t)/dt$ and $B_1(t) = d\,B(t)/dt$. They are solvable, so you can eventually get $A(t), B(t)$. They are most assuredly NOT constants, and that is why you went wrong.

 

[VoteSaxon]

The particular solution should be $y_p(t) = A(t) \cos(2t) + B(t) \sin(2t)$. Substitute that into the DE and equate coefficients of $\sin(2t), \cos(2t)$ on both sides; that will give two coupled 2nd order differential equations for $A(t), B(t)$. These are actually coupled first-order DEs in the functions $A_1(t) = d\,A(t)/dt$ and $B_1(t) = d\,B(t)/dt$. They are solvable, so you can eventually get $A(t), B(t)$. They are most assuredly NOT constants, and that is why you went wrong.

I'm sorry, but this left me in a mess. My professor and his notes have said multiple times that there are only arbitrary constants at the front of the elements in the particular integral. Now you are saying there are new functions introduced instead!

With y''_PI and y_PI in the same function, (y''_PI + 4y_PI = 5sin2t) I am going to have A, B, A₁, B₁, and A₁' and B₁' all as unknowns! It is going to be a mess, surely.

I'm getting a horrible feeling I am out of my depth here ...

Edit: so this is as far as I got... a couple of baffling simultaneous equations ...

(-4B(t) - 4A'(t) + B''(t)) = 5
(-4A(t) + 4B'(t) + A''(t)) = 0

 

[Mark44]

I'm sorry, but this left me in a mess. My professor and his notes have said multiple times that there are only arbitrary constants at the front of the elements in the particular integral. Now you are saying there are new functions introduced instead!

Your professor is correct. The problem is that you have the wrong functions in your particular solution. SteamKing lists one of the functions you'll need in post #2. I would also include Btcos(2t) as part of the particular solution.

With y''_PI and y_PI in the same function, (y''_PI + 4y_PI = 5sin2t) I am going to have A, B, A₁, B₁, and A₁' and B₁' all as unknowns! It is going to be a mess, surely.

I'm getting a horrible feeling I am out of my depth here ...

Edit: so this is as far as I got... a couple of baffling simultaneous equations ...

(-4B(t) - 4A'(t) + B''(t)) = 5
(-4A(t) + 4B'(t) + A''(t)) = 0

 

[VoteSaxon]

Your professor is correct. The problem is that you have the wrong functions in your particular solution. SteamKing lists one of the functions you'll need in post #2. I would also include Btcos(2t) as part of the particular solution.

so... the particular integral is: y_PI = Bsin(2t) + Btcos(2t) ?

 

[Mark44]

so... the particular integral is: y_PI = Bsin(2t) + Btcos(2t) ?

No. The sin(2t) part is one of two basic solutions to the homogeneous equation, y'' + 4y = 0. If {sin(2t), cos(2t)} constitutes a basic set of linearly independent solutions to y'' + 4y = 0, what might you try as a basic set of independent solutions to y'' + 4y = 5sin(2t)?

BTW, I much prefer the term "solution" over "integral" in this context.

 

[Ray Vickson]

I'm sorry, but this left me in a mess. My professor and his notes have said multiple times that there are only arbitrary constants at the front of the elements in the particular integral. Now you are saying there are new functions introduced instead!

With y''_PI and y_PI in the same function, (y''_PI + 4y_PI = 5sin2t) I am going to have A, B, A₁, B₁, and A₁' and B₁' all as unknowns! It is going to be a mess, surely.

I'm getting a horrible feeling I am out of my depth here ...

Edit: so this is as far as I got... a couple of baffling simultaneous equations ...

(-4B(t) - 4A'(t) + B''(t)) = 5
(-4A(t) + 4B'(t) + A''(t)) = 0

The method I used (with non-constant coefficients) is well-known, and is called the "Method of Variation of Parameters". It has been used widely for many decades. See, eg., http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx .

BTW: I do not get your two $A,B$ DEs; I get a system containing only $A', A''$ and $B', B''$ (with no $A,B$); that means we have a first-order system in the derivatives $A_1 = A'$ and $B_1 = B'$, just as I claimed.

However, never mind that: just modify your original method, using the trial form
$$y_p(t) = (A_0 + A_1 t) \sin(2t) + (B_0 + B_1 t) \cos(2t)$$

 

[Mark44]

@VoteSaxon, you might have noticed that the people helping in this thread are giving hints that don't seem connected. That's because all three of us are approaching the problem from a different perspective. SteamKing has provided an "educated guess" particular solution, Ray Vickson is using a technique called "variation of parameters," and I am using a technique that is called "solution by annihilators", with most of my thinking unstated, as you might not be ready for the full explanation. Rest assured that any of the techniques you use will provide a particular solution to your equation, y'' + 4y = 5sin(2t). Once you have a particular solution, it's pretty straightforward getting the general solution to the nonhomogeneous equation.