Question

\[2y^{\prime\prime}+5y^\prime+2y=x^2\]

Answer 1

\[2y_c^{\prime\prime}+5y_c^\prime+2y_c=0\]
\[2m^2+5m+2=0\]\[\Delta=5^2-4(2)(2)=9\]\[(m+2)(2m+1)=0\]\[m_{1,2}=-2,-\frac12\]\[y_c=Ae^{-2x}+Be^{-\frac x2}\]

\[2y_p^{\prime\prime}+5y_p^\prime+2y_p=x^2\]
\[y_p=Cx^2+Dx+E;\quad y_p^\prime=2Cx+D;\qquad y_p^{\prime\prime}=2C\]\[2(2C)+5(2Cx+D)+2(Cx^2+Dx+E)=x^2\]\[(2C)x^2+(10C+2D)x+(4C+5D+2E)=x^2\]\[2C=1:\qquad 10C+2D=0;\qquad 4C+5D+2E=0\]\[C=\frac12 ;\qquad D=-\frac52;\qquad 4\frac12-5\frac52=-2E\]\[-2E=-\frac{21}2;\qquad E=5\frac{1}4\]\[y_p=\frac12{x^2}-\frac52x+5\frac{1}4\]

\[y=y_c+y_p\]
\[y(x)=Ae^{-2x}+Be^{-\frac x2}+\frac12{x^2}-\frac52x+5\frac{1}4\]

Answer 2

ops sorry ive but up the wrong question, i think i have done this all right.

Answer 3

The is the question i ment to ask
\[y^{\prime\prime}+8y^\prime+16y=25e^x\]

Answer 4

\[y_c^{\prime\prime}+8y_c^\prime+16y_c=0\]
\[m^2+8m+16=0\]\[\Delta=(8)^2-4(1)(16)=0\]\[(m+4)^2=0\]\[m=-4\]
\[y_c=(Ax+B)e^{-4x}\]

Answer 5

\[y_p^{\prime\prime}+8y_p^\prime+16y_p=25e^x\]

\[y_p=Ae^{rx};\qquad y_p^\prime=rAe^{rx};\qquad y_p^{\prime\prime}=r^2Ae^{rx}\]\[(r^2Ae^{rx})+8(rAe^{rx})+16(Ae^{rx})=25e^x\]\[\left(r^2+8r+16\right)Ae^{rx}=25e^x\]\[\left(r^2+8r+16\right)Ae^{r}=25\]\[16Ae^r=25\]\[r=1;\qquad A=1\]

\[y_p=e^{x}\]

\[y(x)=(Ax+B)e^{-4x}+e^x\]

Answer 6

im not really sure what i did to get the constants ,
it kinda makes sense now i have them but \[\left(r^2+8r+16\right)Ae^{r}=25\]
how do we find A,r

Answer 7

wouldn't A= 25/16 when r=0, or something

Answer 8

equating co-efficients or some such

Answer 9

How can you cancel this:

\[e^{rx} = e^x\]

Answer 10

i was suing an \(r\) for \(\alpha\)

Answer 11

havent i just timesed by \(e^{-x}\) @waterineyes

Answer 12

oh i see the mistake now

Answer 13

take a look
http://www.efunda.com/math/ode/linearode_undeterminedcoeff.cfm

Answer 14

What you said I did not catch that..

Answer 15

from here\[\left(r^2+8r+16\right)Ae^{rx}=25e^x\]i should have
\[\left(r^2+8r+16\right)Ae^{(r-1)x}=25\]

Answer 16

\[A=25\frac{e^{(1-r)x}}{(r-4)^2}\]

Answer 17

mm confused

Answer 18

Now you are right...

Answer 19

what is the next step?

Answer 20

actually the particular solution for 25e^x is Ae^x\[y_p=Ae^x\]
\[y_p'=Ae^x\]
\[y_p''=Ae^x\]
\[25Ae^x = 25e^x\]
\[A=1\]
\[y=c_1e^{-4x}+c_2xe^{-4x}+e^x\]

Answer 21

If I assume the particular solution as:

\[y_p = ae^x\]

\[y_p' = ae^x\]

\[y_p'' = ae^x\]

Substitute this:

\[ae^x + 8ae^x + 16ae^x = 25e^x\]

\[25ae^x = 25e^x\]

So, a = 1..

\[y_p = e^x\]

\[y(x)=(Ax+B)e^{−4x}+e^x\]

Answer 22

how come you can assume \(Ae^{x}\) as apposed to \(Ae^{rx}\)

Answer 23

See your right hand side..

Answer 24

We assume the particular solution according to the right hand side given so that we can equate and find the value of constants..

Answer 25

ah
r=1
because it has to equal e^{1x}

Answer 26

yes...

Answer 27

that is making some sense

Answer 28

in mukushla's table \(\gamma=1\)

Answer 29

Always choose the particular solution in a way which will make it easy to equate coefficients..

Answer 30

i get it now
thanks everyone

Answer 31

Welcome dear..

Answer 32

\[y_p^{\prime\prime}+8y_p^\prime+16y_p=25e^x\]
\[y_p=Ae^{x};\qquad y_p^\prime=Ae^{x};\qquad y_p^{\prime\prime}=Ae^{x}\]\[(Ae^{x})+8(Ae^{x})+16(Ae^{x})=25e^x\]\[\left(1+8+16\right)Ae^{x}=25e^x\]\[25A=25\]\[A=1\]
\[y(x)=(Ax+B)e^{-4x}+e^x\]

Answer 33

Yes this one is very true..

Answer 34

\[\color{red}\checkmark\]

Answer 35

\[\Huge \color{green} { {\cancel{\heartsuit}}}\]

Answer 36

what does \[ \color{green} { {\cancel{\heartsuit}}}\]mean

Answer 37

I want to make arrow in that heart..

Answer 38

It is simply a sign of smile and happiness...

Answer 39

how do you makes those symbols o.o
also how to write fractions
\[a/b\]

Answer 40

\frac{a}{b} \]

Answer 41

http://en.wikibooks.org/wiki/LaTeX/Mathematics

Answer 42

O.o this is new
ty vm