# Solve the differential equation by method of undetermined coefficient

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

• 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. – Mario SOUPER Mar 24 '18 at 12:13
• I did not get you. would you help to answer this one ? – M. A. SARKAR Mar 24 '18 at 12:20
• 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. – 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.... $$y_p=z_pe^t +u_pe^{2t}$$

• solving the given two equations , we get two particular integral. then what would be the particular integral of the original DE – M. A. SARKAR Mar 24 '18 at 16:01
• @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 – Isham Mar 24 '18 at 16:03
• @yourmath I added some lines I hope it's more clear now.. – Isham Mar 24 '18 at 16:26
• yes , it is clear now. Thank you so much – M. A. SARKAR Mar 24 '18 at 16:39