Question

y^(4)-4y" =t^2 +e^t
I don't know how to set up partial solution for t^2,
Please help

Answer 1

@SithsAndGiggles

Answer 2

Undetermined coefficients?

Answer 3

yes

Answer 4

I'm sure you have the homogeneous solution, so I'll get right to the nonhomogeneous part. As a guess, I'd try \(y_p=At^4+Bt^3+Ct^2+Dt+E\). Then,
\[\begin{cases}
y_p'=4At^3+3Bt^2+2Ct+D\\
\color{red}{y_p''=12At^2+6Bt+2C}\\
y_p^{(3)}=24At+6B\\
\color{red}{y_p^{(4)}=24A}
\end{cases}\]
I haven't worked it out completely, but do you see why I chose this \(y_p\)?

Answer 5

I see you choose the polynomial degree 4. but don't know why

Answer 6

Look at the derivatives of \(y_p\) and the given equation. The first and third derivatives don't matter, since you have \(y^{(4)}-4y''\), so you want to choose an appropriate \(y_p\) such that its derivatives work well with the equation. Does that make sense?

Answer 7

yes, friend

Answer 8

The only foreseeable problem would be a conflict with one of the solutions from the homogeneous set, since you have a repeated root...

Answer 9

@Loser66 the key is that we need \(y''\) that produces a \(t^2,e^t\) terms. It makes sense, then, to start with \(y=A+Bt+Ct^2+Dt^3+Et^4+Fe^t\) and thus \(y''=2C+6Dt+12Et^2+Fe^t\) and \(y^{(4)}=24E+Fe^t\) hence:$$24E+Fe^t-4(2C+6Dt+12Et^2+Fe^t)=t^2+e^t\\24E+Fe^t=8C+24Dt+(48E+1)t^2+(4F+1)e^t$$Hence we observe the following relationship by equating coefficients:$$24E=8C\\D=0\\48E+1=0\\4F+1=F$$solving we get \(C=-1/16,D=0,E=-1/48,F=-1/3\) and note \(A,B\) are free and undetermined by our equation (these correspond to our homogeneous solutions). Our solution is thus:$$y=A+Bt-\frac1{16}t^2-\frac1{48}t^4-\frac13e^t$$

Answer 10

note this isn't the totally general solution, as there are two other linearly independent solutions to our homogeneous equation we have not yet considered

Answer 11

@oldrin.bataku, does the \(A+Bt\) portion of the solution take care of the repeated 0 from the characteristic equation?

Answer 12

@oldrin.bataku @SithsAndGiggles got it, thanks a lot. The net was so bad. I couldn't access until now.

Answer 13

@SithsAndGiggles exactly correct -- that's why they still lack determined coefficients :-p the other two roots are not handled above, however

Answer 14

@oldrin.bataku I have another question from other one.
if y"' +y' = tan t -pi/2<t<pi/2
what does the condition of t affect the partial solution?

Answer 15

@Loser66 it just guarantees all of the coefficient functions in our linear ordinary differential equation are continuous so we're ensured the existence of a unique solution