Question

For the following DE, check that \(y_1\) is a solution of the corresponding homogeneous equation and then calculate the general solution by the reduction of order method.

\[y^{\prime\prime} -4y^\prime +3y=3x-4;\qquad y_1 =e^x\]

Answer 1

\[y^{\prime\prime} -4y^\prime +3y=3x-4;\qquad y_1 =e^x\]
\[\qquad\qquad\qquad\qquad\qquad\qquad y_1^\prime =e^x\]\[\qquad\qquad\qquad\qquad\qquad\qquad y_1^{\prime\prime} =e^x\]\[y_1^{\prime\prime} -4y_1^\prime +3y_1=0\]\[\left(e^x\right) -4\left(e^x\right)+3\left(e^x\right) =0\]\[\left(1-4+3\right)e^x=0\]%3x-4\]\[\therefore\qquad y_1= e^x\qquad\text{is a solution to the homogenous equation}\]\[y_2=uy_1=ue^x\]\[y_2^\prime=u^\prime e^x+ue^x=(u^\prime+u)e^x\]\[y_2^{\prime\prime}=(u^{\prime\prime}+u^\prime)e^x+(u^\prime+u)e^x\]\[y_2^{\prime\prime}=(u^{\prime\prime}+2u^\prime+u)e^x\]\[y_2^{\prime\prime} -4y_2^\prime +3y_2=3x-4\]\[\left((u^{\prime\prime}+2u^\prime+u)e^x\right) -4\left((u^\prime+u)e^x\right) +3\left(ue^x\right)=3x-4\]\[\left(u^{\prime\prime}+2u^\prime+u -4u^\prime-4u +3u\right)e^x=3x-4\]\[\left(u^{\prime\prime}-2u^\prime\right)e^x=3x-4\]\[u^{\prime\prime}-2u^\prime=\left(3x-4\right)e^{-x}\]\[\text{let } p=u^\prime\]\[p^\prime=u^{\prime\prime}\]\[p^\prime-2p=\left(3x-4\right)e^{-x}\]\[\mu(x)=e^{\int-2\text dx}=e^{-2x}\]\[\left(pe^{-2x}\right)^\prime=\left(3x-4\right)e^{-3x}\]\[pe^{-2x}=\int\left(3x-4\right)e^{-3x}\text dx\]\[pe^{-2x}=\int 3e^{-3x}x\text dx-4\int e^{-3x}\text dx\]\[pe^{-2x}=\left[-x{e^{-3x}}-\int -e^{-3x}\text dx\right]+\frac43e^{-3x}\]\[pe^{-2x}=-{xe^{-3x}}-\frac {e^{-3x}}3+\frac43e^{-3x}\]\[pe^{-2x}=-{xe^{-3x}}+e^{-3x}\]\[pe^{-2x}=(1-x)e^{-3x}\]\[p=(1-x)e^{-x}\]\[u^\prime=(1-x)e^{-x}\]\[u=\int(1-x)e^{-x}\cdot\text dx\]\[u=\int e^{-x}\cdot\text dx-\int xe^{-x}\cdot\text dx\]\[u=-e^x-\left[-xe^x-\int -e^{-x}\text dx\right]\]\[u=-e^{-x}+xe^x+e^{-x}\]\[u=xe^{-x}\]\[y_2=uy_1=xe^{-x}e^x=x\]\[y=ay_1+by_2\]

Answer 2

however the answer in the back of the book is
\[y=ae^{x}+be^{3x}+x\]

so i have made some mistakes/ left stuff out

how do i incorporate the not homogenous part of the equation,

Answer 3

hi UnkleRhaukus

for Nonhomogeneous Differential Equations like
\[y''+p(x) y'+q(x) y=r(x)\]
the answer is
\[y(x)=y _{c}+y _{p} \]

Firstly, the 'yc' is the general solution to the homogeneous version of the equation. So you just write down the equation but with ZERO on the right-hand side, use the methods of the previous informations to find the solution, and that's your Complementary Function.

The second part, 'yp' the particular integral, does involve the r(x) on the right-hand side. In fact, by looking at r(x), you have to guess what would be a likely form for the particular integral, and then try it in the equation.

Answer 4

hi mukushla
what kind of guess to make ,

Answer 5

so\[y_c=ay_1+by_2\]\[y_c=ae^x+bx\]\[y=y_c+y_p\]\[y_p=?\]

Answer 6

e^3x?

Answer 7

firstly for finding yc u must drop the (3x-4)

Answer 8

like this?
\[y_c^{\prime\prime} -4y_c^\prime +3y_c=3x-4\]
\[\left(D^2-4D+3\right)y_c=3x-4\]\[y_c=\frac{3x-4}{D^2-4D+3}\]

Answer 9

oh thats thats a different method to find \(y_p\) ;not right here.

Answer 10

\[y''-4y'+3y=0\]is the homogeneous part

Answer 11

yes...

Answer 12

then you make a guess for the particular solution based on the RHS

Answer 13

this method is undetermined coefficients
guess that the particular solution is of the form\[y_c=Ax+B\]then plug that into the original DE and find the val;ues of A and B

Answer 14

...or you could use another method: variation of parameters

Answer 15

which method will take the most lines of working

Answer 16

oh I did neglect the fact that we may have to alter our guess for the particular depending on the solution to the complimentary, so you should find that first

Answer 17

probably undetermined coefficients

Answer 18

but first we should really get the homogeneous part

Answer 19

what have i done already?

Answer 20

you must solve\[y''-4y'+3y=0\]you solved it with the 3x-4 part left in, which is wrong

Answer 21

oh yeah,

Answer 22

you should be able to solve this pretty quick...

Answer 23

\[y_2^{\prime\prime} -4y_2^\prime +3y_2=0\]\[\left((u^{\prime\prime}+2u^\prime+u)e^x\right) -4\left((u^\prime+u)e^x\right) +3\left(ue^x\right)=0\]\[\left(u^{\prime\prime}+2u^\prime+u -4u^\prime-4u +3u\right)e^x=0\]\[\left(u^{\prime\prime}-2u^\prime\right)e^x=0\]

Answer 24

\[u^{\prime\prime}-2u^\prime=0\]

Answer 25

\[\text{let } p=u^\prime\]\[p^\prime=u^{\prime\prime}\]\[p^\prime-2p=0\]

Answer 26

oh my bad, I didn't read that you had to use reduction of order...

Answer 27

\[\mu(x)=e^{\int-2\text dx}=e^{-2x}\]\[\left(pe^{-2x}\right)^\prime=0\]\[pe^{-2x}=\int0\cdot\text dx\]...

Answer 28

should i stop now,

Answer 29

give me a minute to review the process and see if I can't still help...

Answer 30

ok,

Answer 31

yeah, sorry I'm not sure how to do reduction of order with non-homogeneous DE's
I could solve it, but not with that method.

Answer 32

ah, ok thanks for trying Turing test .

Answer 33

good luck Unk

Answer 34

what are you trying to do unkle

Answer 35

i can help i just took a test on this

Answer 36

\[y_2=y_1\int{}{}\frac{e^{-\int{}{}[p(x)dx}}{y_1^2}\]

Answer 37

then use

Answer 38

As you can see this is a pain in the butt so if you can use the reduction formula and the anti derivative comes out easy to do, i'd stick with it. If the antiderivative is something crazy. I'd do it the long way

Answer 39

where did you get that reduction formula @Outkast3r09

Answer 40

@UnkleRhaukus ... let me grab my book

Answer 41

looks like an integrating factor in the numerator

Answer 42

http://books.google.com/books?id=BnArjLNjXuYC&pg=PA150&lpg=PA150&dq=y''-4y'%2B4y%3D0&source=bl&ots=_dUScasRRn&sig=7vXqhtP53Mx2g3LnpHBxRSfgtUU&hl=en&sa=X&ei=YLXwT8GTFaqZ0QGy28z7Ag&sqi=2&ved=0CF4Q6AEwBQ#v=onepage&q=y''-4y'%2B4y%3D0&f=false

Answer 43

page 131 in the middle starts it... then i'll print the second page

Answer 44

the reduction formula comes from seperation of variablees