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Find the general solution?

Find the general solution of u"+2u'+u=25 sin(t).

r^2+2r+1=(r+1)^2

r=-1

u=A sin(t)+B cos(t)

25 sin(t)= -A sin(t)-B cos(t)+2(A cos(t)-B sin(t))+A sin(t)+B cos(t)

= -A sin(t)-B cos(t)+2A cos(t)-2B sin(t)+A sin(t)+B cos(t)

=2A cos(t)-2B sin(t)=A cos(t)-B sin(t)

Now what? Help me step by step.

Comments

-2B is the coefficient of cos(t) on the LHS. (I had a typo in 2A-2B.)

2A is the coefficient of sin(t) on the LHS.

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3 Answers

The homogeneous solution is,

u = C1e^(-t) + C2 t e^(-t)

The steady state solution is,

u* = A sin(t)+B cos(t)

=> u*"+2u*'+u*=25 sin(t)

-Asin(t) - Bcos(t) + 2Acos(t) - 2Bsin(t) + Asin(t) + Bcos(t) = 25 sin(t)

Compare coefficients,

-2B = 25 {edited}

2A = 0
 => A = 0, B = -25/2

So, the general solution is

y = u+u* = C1e^(-t) + C2 t e^(-t) - (25/2)cos(t)

        Ok, so this is a second order linear ordinary differential equation. An "n-th" order general form looks like:

an·u(n)(t) + an-1·u(n-1)(t) + ... + a2·u(2) (t)+ a1·u(1)(t) = F(t)

        Theory concludes that the general solution to the above is the sum between the general solution to the corresponding homogeneous equation:

an·u(n)(t) + an-1·u(n-1)(t) + ... + a2·u(2) (t)+ a1·u(1)(t) = 0

and any particular solution to the original (non-homogeneous) equation.

 

        In your case here you first must find the solution to u" + 2u' + u = 0, which you tried and actually succeeded, by attempting u(t) = er·t as a solution ... and derived that parameter "r" must satisfy   r+ 2·r + 1 = 0, in order for   u(t) = er·t   to satisfy   u" + 2u' + u = 0   .... so you concluded that   r= -1,   which IS GOOD, it means that your general solution to the homogeneous equation (u"+2u'+u=0) must include a term of the form a1e-t.

        The question is -- have you exhausted ALL POSSIBILITIES for detecting all components of the general solution for your homogeneous equation ?  How can you be sure you found the (all-encompassing) general solution ?

Let's start from the fact that a solution of the form  u = er·t  for   u"+2u'+u = 0  actually yields the polynomial equation of second order in the parameter "r" :

r+ 2·r + 1 = (r - 1)2 = 0

which in general may have 2 distinct solutions, and solving it as   (r - 1)(r - 1) = 0   does actually yield 2 (this time identical) solutions:   r = -1   and   r = -1.

        To generate a general solution for   u"+2u'+u = 0   one should add the exponential functions having parameter "r" take BOTH root values (r1 = -1, r2 = -1) as  u(t) = a1e-t + a2e-t. But such a solution reduces to   u(t) = (a1+a2)·e-t   indicating that its 2 component terms -- which should have been independent of each other -- are actually not so: one can be expressed as the other × a constant.   So is there a way to convert one term as to have a new  independent pair of terms ???

        Let us note that searching for "r" such that    u = er·t   satisfies the general case   A·u" + B·u' + C·u = 0   will yield "r" values such that   A·r2 + B·r + C = 0   specifically   r1,2 = -(B/2A) ± Δ,   where   Δ= [ √(B2 - 4·A·C) ] / 2A.

        The exceptional chance here is that in our case   B2 = 4·A·C   ↔   Δ = 0   ↔   r1 = r2 = -(B/2A) and this will help because ...

... if now tried a different solution like   u = t·er·t   for   A·u" + B·u' + C·u = 0   then parameter "r" would have to satisfy:    t·(A·r2 + B·r + C) + 2Ar + B = 0   which is an identity if and only if   { r = -(B/2A)   and   r = -(B/2A) ± Δ }   ↔   { r = -(B/2A)   and   Δ = 0 }   ↔   { r = -(B/2A)   and   B2 = 4·A·C }.   Since in our case this condition is actually fulfilled it follows that not only   u1 = a1·er·t   but also    u2 = a2·t·er·t   is a solution for any equation   A·u" + B·u' + C·u = 0   in which coefficients satisfy   B2 = 4·A·C    (which includes our equation).

        Last but not least, solutions   "u1"   and   "u2"   are independent of each other (one cannot be written as a sum with fixed coefficients from the other).

Thus the general solution to   u" + 2·u' + u = 0   is   u(t) = a1·e-t + a2·t·e-t   .

        Now as for a particular solution to   u" + 2u' + u = 25·sin(t)   it suffices to note that   cos(t),   -sin(t),   -cos(t),   sin(t),   cos(t),  ....  are each the derivative of the prior — so for any combination  u = A·cos(t) + B·sin(t)   we'd have   (u" + u) = 0   that simplifies the original equation to   2·u' = 25·sin(t),   where upon replacing   u' = -A·sin(t) + B·cos(t)   we check that we'd have to have   B = 0   and   A = -25/2.

This is a linear differential equation which is in physics is classified as an equation for linear oscillations with friction (described by the second term 2u') and external force (25sin(t)). It is known that the general solution of the equation is the sum of general solution of the homogeneous equation

                                                                u" + 2u" + u = 0

and one partial solution of the inhomogeneous equation.

If you introduce a new function y = u+u' then the orifinal equation can be rewritten in the following form:

                                                              y'+y = 25sin(t)                                     (1)

To find a patial solution of the equation we can put

                                                            y = A sin(t) + B cos(t)                            (2)

which gives us

                                                           y' = A cos(t) - B sin (t)                            (3)

To make the equation (1) true we have to put y and y' from (2) and (3) into (1) and require

                                                              A + B = 0

                                                              A - B = 25

which gives us  2A = 25 and A = - B. Thus, we have  A = 12.5 and B = - 12.5. Then

                                                   y = 12.5 (sin (t) - cos (t)).                            (4)

If you look back at the equation (1) again you will notice that because cos (t) = sin' (t), then it is easy to

notice that the partial solution (u1) of original euqation can be choosen as

                                                         u1 = -12.5 cos (t)

The next steps are the ones already described by my colleagues above.

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