# Having problem with the particular equation of 2nd order non-homogeneous differential equations

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

• @LutzL no the correct one is the latter \${\rm e} Commented Jul 23, 2019 at 12:26
• @nmasanta its$$e^t not \, \ell^t$$ Commented Jul 26, 2019 at 7:56

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.

• I appreciate your response 😊 but a small elaboration of the power series (frobenius method) would do Commented Jul 23, 2019 at 23:55

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 $$e^{iax}\frac{1}{f(D+ia)} x^n$$

$$5.$$ $$\frac{1}{f(D)} x^{n} (\cos ax + i\sin ax) = \frac{1}{f(D)} x^n e^{iax}=e^{iax}\frac{1}{f(D+ia)} x^n$$

• Which method is this? Undetermined coefficient or variation of constant Commented Jul 27, 2019 at 1:33
• Second order ordinary differential equation with constant coefficient C.F + P.I. method Commented Jul 27, 2019 at 1:43
• I find this solution of yours totally different. Can you recommend a link for me to use, like a tutorial video Commented Jul 27, 2019 at 1:45
• "Introductory Course in Differential Equations" by Daniel Alexander Murray (Chapter VI) Commented Jul 27, 2019 at 1:55
• Also you can see my solutions on this site regarding "ordinary differential equation". math.stackexchange.com/… Commented Jul 27, 2019 at 1:58