Question

Solve by variation of parameters:
y'' + y = sinx

Y'' + Y = 0
\[\lambda = \pm i\]
So, \[y_{c} =Acosx+Bsinx\]
How come the solutions to the complementary function are \(y_1 = cosx\) , \(y_2 = sinx\)??

Answer 1

becasue e^(ix) is another way to write some trig functions

Answer 2

Actually, I don't understand how we can separate them.
I know when it has complex roots \(a\pm bi\) for homogeneous equation, we have the solution
\[y=e^{ax}[Acos(bx)+Bsin(bx)]\], but how can we write it into two roots?

Answer 3

distributive property
\[y=Ae^{ax}cos(bx)+Be^{ax}sin(bx)\]
voila!! 2 roots :)

or rather, lets start out with the roots you found\[y_h=c_1e^{ix}+c_2e^{-ix}\]then see if we can work it into thats

Answer 4

For complex roots \(a \pm bi \)\[y = c_{1} e^{(a+bi)x} + c_{2} e^{(a-bi)x}\] \[ y = e^{ax} (c_{1} e^{(bi)x} + c_{2} e^{(-bi)x})\]By \( e^{ix} = cosx + isinx \)

\[ y = e^{ax} (c_{1} (cos(bx) + isin(bx) + c_{2} cos(-bx) + isin(-bx)) \]\[ y = e^{ax} (c_{1} (cos(bx) + isin(bx)) + c_{2} (cos(bx) - isin(bx))) \]\[y = e^{ax} (c_{1} cos(bx) + c_{2} cos(bx) + c_{1}isin(bx)- c_{2}isin(bx)) \]\[y = e^{ax} [(c_{1} + c_{2}) cos(bx) + (c_{1}- c_{2})isin(bx)] \]\[y = e^{ax} (A cos(bx) + Bsin(bx)) \]
In this case, a = 0, b=1. So,
\[y_{c} = (A cosx + Bsinx) \]

Answer 5

But does that mean the two roots to the complementary function(I meant the one I have to use for finding the particular solution) are cosx and sinx?

Answer 6

nice write up :)

and yes, sinx and cosx are the solutions to the homogenous, some books say compliment, and are used to determine the particular solution

Answer 7

So, the key idea is when I see \(y_{c} = c_1 f_{1}(x)+c_2f_{2}(x)\), then the two solutions I need for finding the particular solution are \(f_{1}(x)\) and \(f_{2}(x)\)?

Answer 8

correct, are you going to use the wronskian? or go at it in the longhand mathod?

Answer 9

the 2nd D of cosx = -cosx
the 2nd D of sinx = -sinx

y'' + y = 0 in both cases just as a double check for the homogenous part

Answer 10

Wronskian? I only know I can use it to determine if the equation is linearly dependent..

Answer 11

the Wronskian works like the cramer rule, or a cross product

Answer 12

Oh! That's another trouble, which I'm going to post it as a new question..
I don't understand how it works!

Answer 13

Should I post my problem in this post? Actually, it's the same question (equation), but I have two problems concerning this equation. The first one is (somehow) solved now.

Answer 14

\[\begin{vmatrix}
W_x&W&W_y\\
f_1&0&f_2\\
f'_1&g&f'_2
\end{vmatrix}\]

Answer 15

\[Wx=0f'_2-gf_2\]
\[Wy=gf_1-0f'_1\]
\[W=f_1f'_2-f_2f'_1\]

Answer 16

this is just a shortcut of the longhand version becasue the long hand ends up using a cramer rule

Answer 17

let A and B be 2 unknown functions of x: A(x) + B(x)

y = A(x)cos(x) + B(x)sin(x)
y' =A'(x)cos(x)-A(x)sin(x) + B'(x)sin(x) +B(x)cos(x)

let A'(x)cos(x) + B'(x)sin(x) represent the homogenous solution; so it equals 0

y' =-A(x)sin(x)+B(x)cos(x)
y'' =-A'(x)sin(x)-A(x)cos(x)+B'(x)cos(x)-B(x)sin(x)

y'' + y = sin(x)

y = A(x)cos(x) +B(x)sin(x)
y'' =-A'(x)sin(x)-A(x)cos(x)+B'(x)cos(x)-B(x)sin(x)
---------------------------------------------\
sin(x) =-A'(x)sin(x) +B'(x)cos(x)

solve the system

A' cos(x) + B'sin(x) = 0
-A' sin(x) + B'cos(x) = sin(x)

Answer 18

Oh!!! Thanks! Actually, I've solved it by those two equations already. But then my teacher said something I don't understand :(

Answer 19

i hope thats a reply your typing and not a lag on the site :)

Answer 20

\[y'' + y =sinx\]\[Y''+Y=0\]\[Y=c_1cosx+c_2sinx\]\[y_p = u_1cosx+u_2sinx\]\[u_1' cosx+u_2'sinx=0 -(1)\]\[u_1' (-sinx)+u_2'cosx=sinx-(2)\]From (1), we get \[u_1' = -\frac{u_2'sinx}{cosx}-(3)\]Sub (3) into (2)
\[( -\frac{u_2'sinx}{cosx}) (-sinx)+u_2'cosx=sinx\]\[u_2' = sinxcosx\]\[u_2 = \frac{1}{4}cos2x\]
\[u_1' = -sin^2x\]\[u_1=-\frac{1}{2}\int(1-cos2x)dx=-\frac{1}{2}x +\frac{1}{4}sin2x\]
\[y_p = (-\frac{1}{2}x +\frac{1}{4}sin2x)cosx+\frac{1}{4}cos2x(sinx)\]\[=-\frac{1}{2}xcosx + \frac{1}{2}(1-sin^2x)(sinx)+\frac{1}{4}(1-2sin^2x)sinx\]\[= -\frac{1}{2}xcosx+\frac{1}{4}sinx\]So, \[y=c_1cosx+c_2sinx -\frac{1}{2}xcosx+\frac{1}{4}sinx\]\[y=c_1cosx+c_2sinx -\frac{1}{2}xcosx\]

Answer 21

That's how I solve this question yesterday/today. But my teacher said something confusing today. It's (somehow) related to undetermined coefficients.

Answer 22

what was the confusing thing?

Answer 23

Here are what I've copied in the lesson today (in latex):
\[y''+y=sinx\]\[\text{Remove or destroy sinx}\]\[D=\frac{d}{dx}\]\[(D^2+1)sinx=0\]\[(D^2+1)[(D^2+1)y] = (D^2+1)sinx\]
^not quite sure of this step\[(D^2+1)^2y = 0\]\[e^{mx}(m^2+1)^2=0\]Don't understand what's going one here
\( => \pm i, \pm i\)
\[sinx, cosx, xcosx, xsinx\]\[y''+y=xcosx\]^Why we choose xcosx, and we can replace 0 by xcosx in the equation?\[(D^2+1)^2xcosx=0\]

Answer 24

I think I got this step: \((D^2+1)[(D^2+1)y] = (D^2+1)sinx\)
\[(D^2+1)sinx = 0\]Replace sinx by \((D^2+1)\) on the left and 0 by \((D^2+1)sinx\) on the right to get that step..

Answer 25

*Replace sinx by \((D^2+1)y\)

Answer 26

No, I think I got the concept wrong!

Answer 27

i havent played around enough with using the D method to be sure whats actually going on either

Answer 28

i understand it up to here
\[(D^2+1)sinx=0\]
but then id want to use quadratic methods to solve D^2+1=0; when D\(=\pm\sqrt{-1}\)
which makes me think im not doing it correctly

Answer 29

I just know the fact that \((D^2+1)sinx=0\)
(sin)'' = -sinx
=> (sinx)''+sinx = 0
But I don't quite understand why we have to use this fact :|

Answer 30

It looks like the given equation though.
y'' + y = sinx
=> \((D^2+1)y = sinx\)

Answer 31

http://www.codecogs.com/reference/maths/calculus/differential/the_d_operator.php

this looks readable to me, and im looking over it to see if i can glean any insight from it

Answer 32

So, replace sinx by \((D^2 +1)y\) in \((D^2+1)sinx=0\), and we get \((D^2+1)(D^2+1)y=0\)

Answer 33

Nice! Now, at least I get to the step \((D^2+1)^2y=0\)

Answer 34

Now, from that readable thingy,
Let u = \((D^2+1)y\)
So, we get
\[(D^2+1)u=0\]
u= \(c_1e^{ix}+c_2e^{-ix}\) ?!

Answer 35

I little more explanation on what happens to sin(x)/4, please

Answer 36

it just gets absorbed into the undetermined constant when we factor.
\[c_2\sin(x)+\frac14\sin(x)=(c_2+\frac14)\sin(x)\]
c2 + 1/4, is just another unknown constant.

the solutions have to be linearly independant (unique), so if we have like terms, they are not different from each other and represent the same term.

Answer 37

as opposed to the cos parts:

c1 cos(x) and -1/2x cos(x) are not the same function due to the extra variable 'x'

Answer 38

I don't understand part of the math done where the complex root homogeneous soln general formula is derived from Euler's Formula; I'm busy right now and can't hop on this immediately, but it would be much appreciated if someone could later answer some questions fielded here.