Choosing a Trial Function for Differential Equation Homework

[Calu]

Homework Statement

I have a differential equation: $\ddot{x} -2\dot{x} + 5x= 10 + 13cos(3t)$

Homework Equations

x(t) = x_c + x_p
where x_c is the Complementary Function and x_p is the Particular Integral.

The Attempt at a Solution

I have formed and solved the auxiliary equation:

$m^{2} - 2m + 5 = 0$

$m_{1} = 1 + 2i , m_{2} = 1 - 2i$

How would I go about forming x_c = e^t(Acos2t + Bsin2t) ?

Furthermore, how would I choose a trial function to determine the Particular Integral?

 

[LCKurtz]

Homework Statement

I have a differential equation: $\ddot{x} -2\dot{x} + 5x= 10 + 13cos(t)$

Homework Equations

x(t) = x_c + x_p
where x_c is the Complimentary Function and x_p is the Particular Integral.

That's "complementary".

The Attempt at a Solution

I have formed and solved the auxiliary equation:

$m^{2} - 2m + 5 = 0$

$m_{1} = 1 + 2i , m_{2} = 1 - 2i$

How would I go about forming x_c = e^t(Acos2t + Bsin2t) ?

I don't understand that question. You did just "form" it, didn't you?

Furthermore, how would I choose a trial function to determine the Particular Integral?

Well, since you need a constant and a $\cos t$ after you differentiate your $x_p$ I would suggest something of the form $x_p = A + B\cos t + C\sin t$.

 

[Calu]

That's "complementary".

I give up with English. Fixed the OP.

I don't understand that question. You did just "form" it, didn't you?

We have taught to from what is called an auxiliary function by taking the coeffeicents of $\ddot{x}, \dot{x},$and $x$ and (say a, b, c) and forming the equation

$am^{2} + bm + c$

and then solving this equation to find 2 values of m (m₁ and m₂) which are to then form the basis of the complementary function x_c. The complementary function I've given above is what I have to form given my values m₁ and m₂.

Well, since you need a constant and a $\cos t$ after you differentiate your $x_p$ I would suggest something of the form $x_p = A + B\cos t + C\sin t$.

That makes sense, thanks. In the given answer a constant 3 has been added so that it reads:

$x_p = A + B\cos 3t + C\sin 3t$. Is there any particular reason that has been chosen?

 

[LCKurtz]

I give up with English. Fixed the OP.



We have taught to from what is called an auxiliary function by taking the coeffeicents of $\ddot{x}, \dot{x},$and $x$ and (say a, b, c) and forming the equation

$am^{2} + bm + c$

and then solving this equation to find 2 values of m (m₁ and m₂) which are to then form the basis of the complementary function x_c. The complementary function I've given above is what I have to form given my values m₁ and m₂.



That makes sense, thanks. In the given answer a constant 3 has been added so that it reads:

$x_p = A + B\cos 3t + C\sin 3t$. Is there any particular reason that has been chosen?

Your original post had a $\cos t$ on the right side. Was that a typing error and it was supposed to be $\cos(3t)$? That would be the only reason to use $3t$ in the trig functions for your $x_p$.

 

[Calu]

Ahh yes, my mistake. Thank you.

Do you have any idea about forming the complementary function?

 

[Mark44]

How would I go about forming x_c = e^t(Acos2t + Bsin2t) ?

Furthermore, how would I choose a trial function to determine the Particular Integral?

Do you have any idea about forming the complementary function?

You already have the complementary function (i.e., the solution to the homogeneous problem). Are you asking about how to write a particular solution (not the complementary solution)? (BTW, the p subscript in x_p stands for "particular.")

If not, then I don't understand what you're having trouble with.

 

[Calu]

I don't know how to go from having m₁ and m₂ to making the equation for x_c. The equation in my post has been given to me in the answer, I don't know how to get it myself.

 

[LCKurtz]

I don't know how to go from having m₁ and m₂ to making the equation for x_c. The equation in my post has been given to me in the answer, I don't know how to get it myself.

Your characteristic equation comes by looking for a solution $x=e^{mt}$ and you have found that $m = 1\pm 2i$. That means the general solution of the homogeneous equation is$$
x_c = Ae^{(1+2i)t} + Be^{(1-2i)t}= e^{t}(Ae^{2it} + Be^{-2it})$$Now use the Euler formulas for the complex exponentials in terms of sines and cosines to get it in the {sine,cosine} form.

 

[Calu]

Your characteristic equation comes by looking for a solution $x=e^{mt}$ and you have found that $m = 1\pm 2i$. That means the general solution of the homogeneous equation is$$
x_c = Ae^{(1+2i)t} + Be^{(1-2i)t}= e^{t}(Ae^{2it} + Be^{-2it})$$Now use the Euler formulas for the complex exponentials in terms of sines and cosines to get it in the {sine,cosine} form.

$x_c = Ae^{(1+2i)t} + Be^{(1-2i)t}= e^{t}(Ae^{2it} + Be^{-2it})$

I did this, but found that I get:

$x_c = e^{t}((A + B)cos2t + (A - B)isin2t)$

Can these arbitrary constants be combined to yield something like:

$x_c = e^{t}(Ccos2t + Disin2t)$ for some arbitrary constants C, D ?

 

[HallsofIvy]

$x_c = Ae^{(1+2i)t} + Be^{(1-2i)t}= e^{t}(Ae^{2it} + Be^{-2it})$

I did this, but found that I get:

$x_c = e^{t}((A + B)cos2t + (A - B)isin2t)$

Can these arbitrary constants be combined to yield something like:

$x_c = e^{t}(Ccos2t + Disin2t)$ for some arbitrary constants C, D ?

Yes, of course. Since A and B are arbitrary constants, so are A+ B and A- B.

 

[Calu]

Yes, of course. Since A and B are arbitrary constants, so are A+ B and A- B.

Thanks.

One more question. Is the $i$ term meant to be present here:

$x_c = e^{t}(Ccos2t + Disin2t)$ ?

 

[LCKurtz]

Thanks.

One more question. Is the $i$ term meant to be present here:

$x_c = e^{t}(Ccos2t + Disin2t)$ ?

Include the $i$ in the $D$. While it appears the constants C and D may be complex, if the DE has real coefficients and real boundary conditions, they will come out real. If you left your answer in the $\{e^{2it},e^{-2it}\}$ form the constants would come out complex.

 

[Calu]

Include the $i$ in the $D$. While it appears the constants C and D may be complex, if the DE has real coefficients and real boundary conditions, they will come out real. If you left your answer in the $\{e^{2it},e^{-2it}\}$ form the constants would come out complex.

Oh I see, thanks a lot!