Given these questions to solve on 2nd order non homogeneous equation. I'm having problem of forming the particular solution of it: $$ \frac{d^2x}{dt^2} \,+4x=289te^t\,sin2t $$ $$ 2x\ddot y\,+\,\dot y \,-\,2y=0 $$ For the first equation, I got the homogeneous equation to be: $ y_c=C_1e^{2it} \,+C_2e^{-2it}$ after forming the particular equation $$ (A+Bt) e^t \sin2t + (C+Dt) e^t\cos2t $$ I applied method of undetermined coefficient but I got incorrect answer Any idea of forming a better solution even without using method of undetermined coefficient
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- $\begingroup$ @LutzL no the correct one is the latter ${\rm e} $\endgroup$Michael Umande– Michael Umande2019-07-23 12:26:00 +00:00Commented Jul 23, 2019 at 12:26
- $\begingroup$ @nmasanta its$$e^t not \, \ell^t$$ $\endgroup$Michael Umande– Michael Umande2019-07-26 07:56:02 +00:00Commented Jul 26, 2019 at 7:56
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2 Answers
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1 First equation, undetermined coefficients requires to see that the right side is not in resonance with the left side, thus the standard construct applies $$ x_p(t)=(A+Bt)e^t\sin(2t)+(C+Dt)e^t\cos(2t). $$
For the second equation the power series approach is called Frobenius method. The Euler-Cauchy part of the equation is $2x^2\ddot y+x\dot y=0$, so that the indicial equation $0=2m(m-1)+m=m(2m-1)$ gives basis solutions $1$ and $\sqrt x$, the power series are thus $$ y_1(x)=\sum_ka_kx^k~~\text{ and }~~y_2(t)=\sqrt x\sum_k b_kx^k. $$ Inserting and comparing coefficients should give the coefficient recursion for both.
Setting $y(x)=\sum a_kx^{k+r}$ with $y'(x)=\sum (k+r)a_kx^{k+r-1}$ and $y''(x)=\sum (k+r)(k+r-1)a_kx^{k+r-2}$ gives after comparing coefficients of equal degree terms $$ 2(k+r)(k+r-1)a_k+(k+r)a_k-2a_{k-1}=0\\~\\ (k+r)(k+r-\tfrac12)a_k=a_{k-1} $$ With $a_{-1}=0$ a non-trivial $a_0$ is only obtained for $r=0$ and $r=\frac12$. Then iterate \begin{align} r&=0:& a_k&=\frac{a_{k-1}}{k(k-\frac12)}&\implies~~&a_k=\frac{2^k}{(2k)!}a_0\\ r&=\tfrac12:&a_k&=\frac{a_{k-1}}{k(k+\frac12)}&\implies~~&a_k=\frac{2^k}{(2k+1)!}a_0 \end{align} Now compare with the power series of the hyperbolic functions.
- $\begingroup$ I appreciate your response 😊 but a small elaboration of the power series (frobenius method) would do $\endgroup$Michael Umande– Michael Umande2019-07-23 23:55:34 +00:00Commented Jul 23, 2019 at 23:55
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5 Given differential equation is $$\frac{d^2x}{dt^2} \,+4x=289~t~e^t~\sin2t\implies (D^2+4)x=289~t~e^t~\sin2t\qquad \text{where }\quad D\equiv \frac{d}{dt}$$
Roots of the trial solution are $~\pm 2i~$, so the complementary function (C.F.) is $$a~\cos(2t)+b~\sin(2t)\qquad $$where $~a,~b~$ are independent constants.
Now the particular integral (P.I.) is
P.I. $~=\frac{1}{D^2+4}289~t~e^t~\sin2t$
$~~~~~~~= 289~e^t~\frac{1}{(D+1)^2+4}t~\sin2t$
$~~~~~~~= 289~e^t~\frac{1}{D^2+2D+5}t~\sin2t$
$~~~~~~~= 289~e^t~\left(\text{imaginary part of}\left\{~\frac{1}{D^2+2D+5}t~e^{2it}\right\}\right)$
$~~~~~~~= 289~e^t~\left(\text{imaginary part of}\left\{~e^{2it}~\frac{1}{(D+2i)^2+2(D+2i)+5}t~\right\}\right)$
$~~~~~~~= 289~e^t~\left(\text{imaginary part of}\left\{~e^{2it}~\frac{1}{D^2+(2+4i)D+(1+4i)}t~\right\}\right)$
$~~~~~~~= 289~e^t~\left(\text{imaginary part of}\left\{~\frac{e^{2it}}{1+4i}~\left(1-\frac{2+4i}{1+4i}~D+\cdots\right)t~\right\}\right)$
$~~~~~~~= 289~e^t~\left(\text{imaginary part of}\left\{~\frac{e^{2it}}{1+4i}~\left(t-\frac{2+4i}{1+4i}\right)~\right\}\right)$
$~~~~~~~= 289~e^t~\left(\text{imaginary part of}\left\{~\frac{1-4i}{17}~(\cos~2t+i\sin~2t)~\left(t-\frac{18-4i}{17}\right)~\right\}\right)$
$~~~~~~~= 289~e^t~\left(\text{imaginary part of}\left\{~\frac{1}{17}~[(\cos~2t+4\sin~2t)+i(\sin 2t-4\cos 2t)]~\left(t-\frac{18-4i}{17}\right)~\right\}\right)$
$~~~~~~~= 289~e^t~\left(~\frac{4}{17^2}~(\cos~2t+4\sin~2t)+\frac{1}{17}~(\sin 2t-4\cos 2t)~\left(t-\frac{18}{17}\right)~\right)$
$~~~~~~~= ~e^t~\{~4~(\cos~2t+4\sin~2t)+~(\sin 2t-4\cos 2t)~\left(17~t-18\right)~\}$
So the general solution is
$$x(t)=C.F. + P.I.$$
$$x(t)=a~\cos(2t)+b~\sin(2t)+~e^t~\{~4~(\cos~2t+4\sin~2t)+~(\sin 2t-4\cos 2t)~\left(17~t-18\right)~\}$$where $~a,~b~$ are independent constants.
For the Particular Integral (i.e., P.I.) there are some general rules
$1.$ $\frac{1}{D + a} \phi (x) = e^{-ax}\int e^{ax}\phi(x)$
$2.$ $\frac{1}{f(D)} e^{ax} \phi(x) = e^{ax}\frac{1}{f(D+a)} \phi(x)$
$3.$ $\frac{1}{f(D)} x^{n} \sin ax = $Imaginary part of $e^{iax}\frac{1}{f(D+ia)} x^n$
$4.$ $\frac{1}{f(D)} x^{n} \cos ax = $Real part of $e^{iax}\frac{1}{f(D+ia)} x^n$
$5.$ $\frac{1}{f(D)} x^{n} (\cos ax + i\sin ax) = \frac{1}{f(D)} x^n e^{iax}=e^{iax}\frac{1}{f(D+ia)} x^n$
- $\begingroup$ Which method is this? Undetermined coefficient or variation of constant $\endgroup$Michael Umande– Michael Umande2019-07-27 01:33:19 +00:00Commented Jul 27, 2019 at 1:33
- $\begingroup$ Second order ordinary differential equation with constant coefficient C.F + P.I. method $\endgroup$nmasanta– nmasanta2019-07-27 01:43:17 +00:00Commented Jul 27, 2019 at 1:43
- $\begingroup$ I find this solution of yours totally different. Can you recommend a link for me to use, like a tutorial video $\endgroup$Michael Umande– Michael Umande2019-07-27 01:45:44 +00:00Commented Jul 27, 2019 at 1:45
- $\begingroup$ "Introductory Course in Differential Equations" by Daniel Alexander Murray (Chapter VI) $\endgroup$nmasanta– nmasanta2019-07-27 01:55:34 +00:00Commented Jul 27, 2019 at 1:55
- $\begingroup$ Also you can see my solutions on this site regarding "ordinary differential equation". math.stackexchange.com/… $\endgroup$nmasanta– nmasanta2019-07-27 01:58:01 +00:00Commented Jul 27, 2019 at 1:58