Question

How do i find the particular solution to a second order differential equation \[ay^{\prime\prime}+by^\prime+cy=f(x)\] where \(f(x)\) is a hyperbolic function

Answer 1

can i use undetermined co-efficients or dosen't this work?

Answer 2

You can use that, I think..

Answer 3

\[y^{\prime\prime}-y=\cosh x\]
\[y_c^{\prime\prime}-y_c=0\]\[m^2-1=0\]\[m=\pm1\]
\[y_c=Ae^x+Be^{-x}\]

\[y_p^{\prime\prime}-y+p=\cosh x\]

Answer 4

Why you have added p here??

Answer 5

what form can i assume \(y_p\) is of
\[y_p=C\cosh x+D\sinh x\]?

Answer 6

Yes you are right..

Answer 7

i kinda guessed that , can you link me to a reference ?

Answer 8

What do you mean??

Put:

\[y_p = Ccoshx + Dsinhx\]

Take the derivative and find \(y_p''\)..

Answer 9

\[\huge \color{green}{\frac{d}{dx}sinhx = coshx}\]

\[\huge \color{green}{\frac{d}{dx}coshx = sinhx}\]

Answer 10

@UnkleRhaukus it is because they all are some form of e^(x) and so we use eulers identity. i dont know if thats what your asking...

Answer 11

ill try it, i just wasent sure if i had guessed the correct form, i can seam to find it in any table for particular solution

Answer 12

\[y_p^{\prime\prime}-y+p=\cosh x\]
\[y_p=C\cosh x+D\sinh x;\qquad y^\prime_p=C\sinh x+D\cosh x;\qquad y^{\prime\prime}_p=C\cosh x+D\sinh x\]

Answer 13

what is p here??

Answer 14

ops i dont know, p=0

Answer 15

no wait i meant
\[y_p^{\prime\prime}-y_p=\cosh x\]

Answer 16

I think we can't put \(Acoshx + Dsinhx\), it will become 0 on putting in that DE..

Answer 17

\[C\cosh x+D\sinh x-C\cosh x+D\sinh x=\cosh x\]
\[0=\cosh x\]
so i have made a false assumption,

Answer 18

We can solve the right hand side for coshx..

Answer 19

\[\cosh x\neq0\]

Answer 20

Think about this:
May be it will help you:

\[\huge \color{blue}{coshx = \frac{e^x + e^{-x}}{2}}\]

\[\huge \color{blue}{sinhx = \frac{e^x - e^{-x}}{2}}\]

Answer 21

If we substitute this in place of coshx then we have particular solution is \(e^x\) and \(e^{-x}\) terms..

Getting or not??

Answer 22

yeah so there is not a formula for the particular solution if f(x) is hyperbolic, i just have to break the hyperbolic up , i get this but i was hoping there would be a more elegant method, like an assumable form, i guess not

Answer 23

You could solve it if it is:
\[y'' + y = coshx\]

By putting:

\[y_p = Ccoshx + Dsinhx\]

Answer 24

@waterineyes \[y_p^{\prime\prime}-y_p=\cosh x\]
\[y_p=C\cosh x+D\sinh x;\qquad y^\prime_p=C\sinh x+D\cosh x;\qquad y^{\prime\prime}_p=C\cosh x+D\sinh x\]
\[(C\cosh x+D\sinh x)-(C\cosh x+D\sinh x)=\cosh x\] \(0=\cosh x\) False

this does not work for some reason

Answer 25

I said for:

\[y'' + y = coshx\]

and not:
\[y'' - y = coshx\]..

Answer 26

yeah why is it different with a plus,

Answer 27

The terms will not get cancel..

Answer 28

\[2Ccoshx + 2Dsinhx = coshx\]

Answer 29

yeah i can see that in this case , but in general?

Answer 30

In general you can use that transformation from coshx to exponential..

Answer 31

how can i tell which method to use for other question in the set

Answer 32

Go with exponential..

Answer 33

ok

Answer 34

\[y_p^{\prime\prime}-y_p=\cosh x\]
\[\text{---------}\]
\[y_p^{\prime\prime}-y_p=\frac{e^x}2+\frac {e^{-x}}2\]

\[y_{p_{1}}=Ce^{x};\qquad y^\prime_{p_1}=Ce^{x};\qquad y^{\prime\prime}_{p_1}=Ce^{x}\]
\[Ce^{x}-Ce^{x}-=\frac{e^x}2\]\[0=\frac{e^x}2\] False

\[y_{p_{2}}=Ce^{-x};\qquad y^\prime_{p_2}=-Ce^{-x};\qquad y^{\prime\prime}_{p_2}=Ce^{-x}\]\[Ce^{-x}-Ce^{-x}=\frac{e^{-x}}2\]\[0=\frac{e^{-x}}2\] False


???

Answer 35

I'll again deviate from "guessing" the particular solution, by using the Annihilator Method.
\(y'' - y = \cosh{x}\) Let \(y_{p}\) be the particular solution and \(D = dy/dx\)
\[(D^{2}-1)y_{p} = \frac{e^{x}+e^{-x}}{2}\] To annihilate \(e^{x}\), we see that operating \((D-1)e^{x} = 0\) and to annihilate \(e^{x}\), we see that operating \((D+1)e^{-x}=0\), THerefore operating (D+1)(D-1) on the equation, we get
\[(D^{2}-1)^{2}y_{p}=0 \Rightarrow (D+1)^{2}(D-1)^{2}y_{p} = 0\]
so, \(y_{p}\) is of the form \[y_{p}= ae^{x}+bxe^{x}+ce^{-x}+dxe^{-x}\]As the homogenous solution contains \(e^{x}\) and \(e^{-x}\) as solutions, therefore the particular solution is \(\Large y_{p} = bxe^{x}+dxe^{-x}\)

Answer 36

i got
\[y_p=\frac{3xe^x}{8} + \frac{e^x}{16} +\frac {xe^{-x}}{8}-\frac{e^{-x}{8}\]
somehow ^ doesn't work on my screen


y_p = (3xe^x)/8 + e^x/16 + (xe^-x)/8-e^-x/8

1 bit lemme check wolfram if it's right

Answer 37

\[y_p=\frac{3xe^x}{8} + \frac{e^x}{16} +\frac {xe^{-x}}{8}-\frac{e^{-x}}{8}\]
your just missing a closing brace for the numerator in the last term @nphuongsun93

Answer 38

o lol =.= *new to latex*

but that was wrong.. brb retry

Answer 39

the answer in the back of the book is \[y=\frac x2\sinh x+Ae^x+Be^{-x}\]
so \[y_p=\frac x2\sinh x\]

Answer 40

@UnkleRhaukus , I do get \(y_{p} = (x \sinh{x})/2\) after using what I have done.

Answer 41

\[y'' - y = \frac{e^x}{2} + \frac {e^{-x}}{2}\]
\[y_p1=\frac{e^x}{2}\]
\[y_p = Axe^x\]
\[y_p'=A(x+1)e^x\]
\[y_p''=A(x+2)e^x\]
\[y''-y' = Axe^x+2Ae^x-Axe^x\]
\[2Ae^x=\frac{e^x}{2}\]
\[A = 1/4\]

\[y_p2= Ae^{-x}\]

Answer 42

zz misclick post too soon D: typing the rest

Answer 43

\[y_{p_1}^{\prime\prime}-y_{p_1}=\frac{e^x}2\]How do you know to use this form?\[y_{p_1}=Axe^x\]
Where does the multiplicative x comes from?

Answer 44

ignore the last line
\[y_p2=Bxe^{-x}\]
\[y_p'=-Be^{-x}(x-1)\]
\[y_p''=Be^{-x}(x-2)\]
\[y''-y=e^{-x}/2\]
\[Bxe^{-x}-2Be^{-x}-Bxe^{-x}=e^{-x}/2\]
\[B=-1/4\]

\[y_p = xe^x/4 - xe^{-x}/4\]

Answer 45

you have got the solution i want @nphuongsun93 Thank you and well done

i just dont understand where the x comes from

Answer 46

ok your question
because the homogenous solution is \[c_1e^x+c_2e^{-x}\] and the coefficients for the particular solution happens to be \[Ae^x and Be^{-x}\]
if you leave the same e^x and e^(-x) you will get the wrong answer, so whenever these are the same, just times the coefficients with x.

Answer 47

same thing when you get the same homogenous solution

like \[y''+4y'+4y=0\]
\[r=-2,-2\]
\[y=c_1e^{-2x}+c_2xe^{-2x}\]

sorry i'm not very good with english and wording ;_;

Answer 48

so it is because

\[c_1e^{\color\red x}+c_2e^{\color\red{-x}}\]

\[\color\red {x-x}=0\]?

Answer 49

no you can't add them like that ^
if the homogenous solution is like this
\[c_1e^x+c_2e^{-x} \]
and the particular solution is like \[Ae^x+Be^{-x}\]
the particular solution is gonna get canceled =0 when you uhh.. add them after "derivative" step

Answer 50

i think that is what i mean

Answer 51

oh o: ok want me to give an example?

Answer 52

yes please

Answer 53

\[y''+y=5cosx\]
let's do ^

Answer 54

\[y_p=A\cos x;\qquad y^\prime_p=-A\sin x ;\qquad y_p^{\prime\prime}=-A\cos x\]

Answer 55

\[-A\cos x+A\cos x=5\cos x\]\[0=5\]\error

Answer 56

\[y_p=Ax\cos x;\qquad y^\prime_p=A\cos x-Ax\sin x ;\qquad y_p^{\prime\prime}=-A\sin x-A\sin -Ax\cos x\]

Answer 57

page got reloaded -_-

Answer 58

\[y_p = Axcos(x) + Bxsin(x)\]

Answer 59

whops

Answer 60

\[y_p''=-2Asinx-Acosx+2Bcosx-2Bxsinx\]

Answer 61

\[y_p=Ax\cos x+Bx\sin x\]\[ y^\prime_p=A\cos x-Ax\sin x +B\sin x+Bx\cos x\]\[y_p^{\prime\prime}=-A\sin x-A\sin x -Ax\cos x+B\cos x+B\cos x-Bx\sin x\]

Answer 62

\[y''+y=5cosx\]
\[-2Asinx-2Bcosx=5cosx\]
\[B=-5/2\]

Answer 63

\[y_p = \frac{-5}{2}cosx\]

Answer 64

did i differentiate incorrectly?

Answer 65

oh no i typed wrong again D: ignore that 2 in the 2B x sin(x). (y'')

Answer 66

These are the same kinds of problems as the one we looked at yesterday, where
the driving term on the rhs is a solution of the homogeneous diff equation. So, like yesterday, you need to use a different guess.....

Answer 67

go on,..

Answer 68

For the first one(I'm still catching up here) if the equation where

\[y \prime \prime-y = \cosh(\lambda x)\]

with [\\lambda \neq 1], you'd know the solution was

[\ y= A \cosh x + B \sinh x + \cosh x/ (\lambda^2-1) \]

yes?

Answer 69

i did not know that

Answer 70

The A and B are the homogeneous solution. For the particular solution, you'd guess
that

\[y_p= C \cosh(\lambda x)+ D \sinh(\lambda x)\]

and then when you plugged this in and solved for C and D, you'd find C=0 and D= 1(lambda^2 -1).

Answer 71

yeah in my case \(\lambda\) was 1 so \(D=(\lambda-1)=0\)

Answer 72

Here is the final solution for you to use to check your answer:
\[
\begin{array}{c}
y(x)=-\frac{1}{4} e^{-x}
x+\frac{e^x
x}{4}-\frac{e^{-x}}{8}-\frac{e^x}{8}+e^x c_1+e^{-x}
c_2 \\
\end{array}
\]

Answer 73

Right. Here's the "trick" , the solution we're after isn't the particular soution, but really the full solution. In the full solution, A and B are arbitrary constants that we use to set the boundary conditions(initial values) of the problem. In general, they'll depend om lambda.

So what wer'e going to do is to choose a slightly different guess for the particular solution
by just moving a little of the homogeneous solution into the particular

\[y_p = (\cosh \lambda x - \cosh x)/(\lambda^2 -1) \]

This is still a solution for lambda not equal to 1, because the extra cosh x term
satisfies the homogeneous eq. As lambda goes to 1, this form is indeterminate,
so you have to apply L'hopital's rule to evaluate it, and you get
\[y_p= xsinh x/2\]

Answer 74

Or
\[
\begin{array}{c}
y(x)=-\frac{1}{4} e^{-x}
x+\frac{e^x
x}{4}+e^x c_1+e^{-x}
c_2 \\
\end{array}
\]

Answer 75

or
\[
\begin{array}{c}
y(x)=\frac{1}{2}x \sinh(x) +e^x c_1+e^{-x}
c_2 \\
\end{array}
\]

Answer 76

When choosing the particular solution, you should prevent any part of it to be a solution
of the homogeneous solution. This is achieved by multiplying the initial guess by x or x^2

Answer 77

\[y_p =\lim_{\lambda \rightarrow1} \frac{\cosh \lambda x - \cosh x}{\lambda^2 -1}\rightarrow \frac{0}{0}\]
\[\stackrel{\text{l'H}}=\frac{\frac{\text d}{\text d\lambda}\left(\cosh \lambda x - \cosh x\right)}{\frac{\text d}{\text d\lambda}\left(\lambda^2-1\right)}\]\[=\frac{\lambda\sinh \lambda x}{2\lambda}\]\[=\frac{\sinh\lambda x}{2}\]

Answer 78

im missing an x

Answer 79

You have to find your particular solution as \( A x e^x + B x e^{-x}\) and find A and B

Answer 80

\[\stackrel{\text{l'H}}=\frac{\frac{\text d}{\text d\lambda}\left(\cosh \lambda x - \cosh x\right)}{\frac{\text d}{\text d\lambda}\left(\lambda^2-1\right)}=\frac{x\sinh\lambda x}{2\lambda}\]?

Answer 81

\[y^{\prime\prime}-y=\cosh x\]the complementary solution
\[y_c^{\prime\prime}-y_c=0\]\[m^2-1=0\]\[m=\pm1\]\[y_c=Ae^x+Be^{-x}\]

Answer 82

i see now

\[\lim_{\lambda\rightarrow 1}\frac{\frac{\text d}{\text d\lambda}\left(\cosh \lambda x - \cosh x\right)}{\frac{\text d}{\text d\lambda}\left(\lambda^2-1\right)}\]\[=\lim_{\lambda\rightarrow 1}\frac{x\sinh\lambda x}{2\lambda}\]\[=\frac{x\sinh x }2\]

Answer 83

\[
2 A e^x-2 B e^{-x}=\frac{1}{2}
\left(e^{-x}+e^x\right)\\
A= \frac 1 4 \\
B= -\frac 1 4\\
y_p=x\frac 1 2 \left( \frac {e^x- e^{-x}}{2}\right)=\frac 1 2 x \sinh(x)
\]

Answer 84

Doing it the usual way that I described seems to me a slighter more known approach.

Answer 85

@satellite73 @lalaly

Answer 86

@eliassaab i just dont know how i would tell when the particular solutions would be part of the homogeneous solution

Answer 87

Looking at most of the posts, it seems to me that we are making a very easy problem into a somewhat a bit more difficult.

Answer 88

this is not a very easy problem

Answer 89

Use ricatti solving method

Answer 90

It is very straight forward, believe me.
Since \( cosh(x) = \frac 1 2( e^x + e^{-x})\) the initial guess is
\( A e^x + B e^{-x} \) these are solutions of the homogenous equation. Multiply by x
and you get \( A x e^x + B x e^{-x} \) . Now none of the two in the sum is a solution of the homogeneous equation.

Answer 91

if you see, for example, e^x or cos(x) or f(x) in the homogenous solution, avoid using those again when you are solving for particular solution by multiplying x to those, for example, xe^x or x cos(x) or x f(x)

Answer 92

if no one solve this for you, im back in 5/6 hours. PM me and i'll do it for you. I dont have time now

Use Ricatti's equation

Answer 93

way back at the start, you identified the homogeneous solution as A sinh x +B cosh x, at that point you can look at the fct that appears on the RHS of the DEQ. If one(or a linear combination)
of the the fcts in the homogeneous solution appears, then the simple guess won't work and you need to try one of these other methods.

Answer 94

i think im getting it now!
i have to look at my homogenous/complimentary solution, if my particular solution is in the same phase it will resonate this mean a factor of x

Answer 95

@UnkleRhaukus. What is the verdict: easy or not?

Answer 96

the whole question for me was where does the factor of x come from, now it makes perfect sense... happy

Answer 97

justifying the factor of x was not easy.