Solve the differential equation by ** method of undetermined coefficient** .

$ y''-5y'+6y=e^{t} \cos 2t+e^{2t} (3t+4) \sin t \ $


The auxiliary equation is

$ m^2-5m+6=0 \\ \Rightarrow m=2,3 $

The complementary function is

$ C.F.=c_1 e^{2t}+c_2 e^{3t} \ $ , where $ \ c_1, c_2 \ $ are arbitrary constants

But I can not assume how to construct the particular integral $ \ P.I. \ $

I think the particular integral $ \ P.I.=Ae^{t} \cos 2t +B t^2 e^{2t} (3t+4) \sin t \ $ ,

because $ \ e^{2t} \ $ appears in the complimentary function .

But I am not sure.

Help me out

  • $\begingroup$ Use the superposition principle, solve $y''-5y'+6y=e^t\cos2t$ and $y''-5y'+6y=e^2t(3t+4)\sin t$, first one for a Polynomial of the zero degree, the solution is of the form $y_{p1}=e^t(A(x)\cos 2t+B(x)\sin 2t)$ where $max(deg A,deg B)=0$, and the second one, since 2 is a simple root of the char. eq., same form of the particular solution with max deg = 2 but the inside of the cos/sin is $(t)$ and not $2t$. Then add the three solutions togother for the general solution. $\endgroup$ – Mario SOUPER Mar 24 '18 at 12:13
  • $\begingroup$ I did not get you. would you help to answer this one ? $\endgroup$ – M. A. SARKAR Mar 24 '18 at 12:20
  • 1
    $\begingroup$ Solve the two DEs, one is of the form, $y_{p1}=e^t(A(x)\cos 2t+B(x)\sin 2t)$ where $max(degA,degB)=0$ (A and B are polynomials), $y_{p2}=e^{2t}(C(x)\cos t+D(x)\sin t)$ where $max(degC,degD)=2$ because $2$ is a root of the characteristic equation. (C and D are polynomials). Then add the two solutions together, with the solution of the homogen. equation. You find the polynomials by substituting in the original DE. $\endgroup$ – Mario SOUPER Mar 24 '18 at 12:22

As Mario pointed out you can split the equation into two equation $$ \begin{cases} y''-5y'+6y=e^{t} \cos 2t \\ y''-5y'+6y=e^{2t} (3t+4) \sin t \end{cases} $$ Substitute for the first $y=ze^t$ and for the second $y=ue^{2t}$

Then simplify both equations $$ \begin{cases} z''-z'+2z=\cos 2t \\ u''-u'= (3t+4) \sin t \end{cases} $$ Solve both equations to get the particular solution

Edit for youmath

you get $$ \begin{cases} z=z_h+z_p \\ u=u_h+u_p \end{cases} \implies \begin{cases} y=z_he^t+z_pe^t \\ y=u_he^{2t}+u_pe^{2t} \end{cases} \implies y=e^tz_h+z_pe^t +u_he^{2t}+u_pe^{2t}$$

The homegeneous part you already have it for y but now you have the particular part too.... $$y_p=z_pe^t +u_pe^{2t}$$

  • $\begingroup$ solving the given two equations , we get two particular integral. then what would be the particular integral of the original DE $\endgroup$ – M. A. SARKAR Mar 24 '18 at 16:01
  • $\begingroup$ @yourmath you substitute back ...with the relations you have $y=ze^t \implies z=ye^{-t} $ the same for u We used the superposition principle since both equations are linear as Mario pointed out $\endgroup$ – Isham Mar 24 '18 at 16:03
  • 1
    $\begingroup$ @yourmath I added some lines I hope it's more clear now.. $\endgroup$ – Isham Mar 24 '18 at 16:26
  • $\begingroup$ yes , it is clear now. Thank you so much $\endgroup$ – M. A. SARKAR Mar 24 '18 at 16:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.