Question

y(4)-4y"=t^2 +e^t, please help

Answer 1

I got stuck when calculating the partial solution,
\[Y_{p1}= At^2 +Bt+C\]

Answer 2

the homogeneous solution is \[y_g= C_1+C_2t+C_3e^{2t}+C_4e^{-2t}\]It's ok,

Answer 3

At^2+bt+C+De^t should be your solution

Answer 4

I separate the partial solution into 2 part, you are right with De^t
I just calculate the polynomial part and get stuck when second derivative is a constant.

Answer 5

\[Y_{p1}=At^2 +Bt +C\\Y'_{p1}=2At+B\\Y"_{p1}=2A\\Y"'_{p1}=0\]
then \[-4(2A) = t^2\]it doesn't make sense at all

Answer 6

@bahroom

Answer 7

@oldrin.bataku

Answer 8

@loser66 there is try solving the equation with \(z=y''\)

Answer 9

if so, I have y_p= -1/4t^2-1/8

Answer 10

how to plug back?

Answer 11

$$z''-4z=0$$Let \(z=e^{\lambda t}\):$$\lambda^2e^{\lambda t}-4e^{\lambda t}=0\\e^{\lambda t}(\lambda^2-4)=0\\\lambda =\pm2$$hence \(z_c=C_1e^{2t}+C_2e^{-2t}\) is the general solution for our homogeneous equation. For our particular solution, observe we need to come up with \(t^2,e^t\) terms on our right-hand side so assume a solution of the form \(z_p=A+Bt+Ct^2+De^t\), yielding \(z_p''=2C+De^t\):$$2C+De^t-4(A+Bt+Ct^2+De^t)=t^2+e^t\\2C+De^t=4A+4Bt+(4C+1)t^2+(4D+1)e^t$$so clearly \(C=2A,D=4D+1,B=0,4C+1=0\) hence \(A=-1/8,B=0, C=-1/4,D=-1/3\) and \(z_p=-\frac18-\frac14t^2-\frac13e^t\). Observe, then, that we have the general solution \(z=z_c+z_p=C_1e^{2t}+C_2e^{-2t}-\frac13e^t-\frac14t^2-\frac18\). To determine the general solution for \(y\), observe:$$z=y''\\y=\int\int C_1e^{2t}+C_2e^{-2t}-\frac13e^t-\frac14t^2-\frac18\,dx\,dx=\dots$$

Answer 12

The reason your method failed is because you guessed a nonsense particular solution. Observe that since we only have \(y'',y^{(4)}\) in our equation, they must contain \(t^2,e^t\) terms, yet your given particular solution "guess" gives \(y''=2A\) and \(y''=0\). Instead, you should've tried with \(y_p=A+Bt+Ct^2+Dt^3+Et^4+Fe^t\)... :-) that seemed ugly, though, so I rewrote it as a 'nicer' linear differential equation of lower order in \(z=y''\) so that you can deal with the ugliness later during integration.

Answer 13

how can I know this method? why z = e^lambda t?

Answer 14

This method is just basic substitution and the basic approach to solving inhomogeneous linear differential equations. Assuming \(z=e^{\lambda t}\) lets us find the fundamental solutions of our homogeneous complementary equation \(z''-z=0\). We then used the method of undetermined solutions to find a particular solution.

Answer 15

without it, we still have lambda = +/- 2?

Answer 16

I don't understand your question

Answer 17

let see the equation after replace z = y"
it's z"-4z = t^2 +e^t
homogeneous part is z" -4z=0
characteristic equation is \(\lambda^2-4=0\\\lambda = \pm2\)
which yield \[Z_g= C_1e^{-2t}+C_2e^{2t}\]

Answer 18

Thanks a lot, You are only one person who can give out a neat solution like this. I appreciate your help

Answer 19

@looser66 yes very good :-) that is indeed your general solution for the homogeneous equation \(z''-4z=0\).

Answer 20

You get the exact same solution for \(y^{(4)}-4y''=0\), in fact :-) Let \(y=e^{\lambda t}\)$$\lambda^4e^{\lambda t}-4\lambda^2e^{\lambda t}=0\\\lambda^2e^{\lambda t}(\lambda^2-4)=0\\\lambda=0\,(\text{mult}\,2),\pm2$$Observe we have a repeated root hence \(y_c=C_1+C_2t+C_3e^{-2t}+C_4e^{2t}\).

The solution for \(z\) is in face equivalent, i.e.:$$z=y''\\y=\int\int z\,dx\,dx=\int\int C_1e^{-2t}+C_2e^{2t}\,dx\,dx=\int-\frac12C_1e^{-2t}+\frac12C_2e^{2t}+C_3\,dx\\y=\frac14C_1e^{-2t}+\frac14C_2e^{2t}+C_3t+C_4$$... with an appropriate change in the labeling of constants :p

Answer 21

got you. so, we must do that step? I mean let y = e^lambda t?
and there are some more steps as you did ( take double integral ) to get the final answer?

Answer 22

Ha !! It takes a long time to solve.

Answer 23

well, with the \(z=y''\) substitution yes you must integrate to get back a solution \(y\) :-p but I think it is easier to substitute first than to do it the long way :p

Answer 24

But anyways we do the \(e^{\lambda t}\) steps because it yields our characteristic equation! :p

Answer 25

You could just memorize the form of the characteristic equation but where's the fun in that? :p more fun to derive

Answer 26

NOOOOOO , it's not fun aaatttt aaaalll.

Answer 27

I saw other person has the interesting problem, I tagged you there, come to help him, I want to see, too. PPleease

Answer 28

differential equation, too.