Question

y'''-2y''-2y'+y=0 where y"(0)=1, y(0)=2, y(0)=2 how to get this were having trouble in this assignment please help :(

Answer 1

https://www.wolframalpha.com/input/?i=y%27%27%27-2y%27%27-2y%27%2By%3D0+where+y%22%280%29%3D1%2C+y%280%29%3D2

Answer 2

First, solve the characteristic equation
\[
r^3-2 r^2-2 r+1=0\\
(1 + r) (1 - 3 r + r^2)=0

\]
The roots are
\[
\{r= -1\},\\
\left\{r= \frac{1}{2}
\left(3-\sqrt{5}\right)\right\},
\\\left\{r=
\frac{1}{2}
\left(3+\sqrt{5}\right)\right\}
\]

Answer 3

You find the general solution
\[
y=a e^{\left(\frac{3}{2}-\frac{\sqrt{5}}{2}\right) x}+b
e^{\left(\frac{3}{2}+\frac{\sqrt{5}}{2}\right) x}+c
e^{-x}
\]

Answer 4

You find a, b and c from the initial conditions.

Answer 5

thank you @eliassaab

Answer 6

i'm now confuse how to find those value :(

Answer 7

i need clear solution please im begging for help

Answer 8

sorry if im too slow to understand...

Answer 9

You have three initial conditions and you need three equations, so from each condition get an equation. From
y(0)=2, you get
\[
2=y(0)=a e^{\left(\frac{3}{2}-\frac{\sqrt{5}}{2}\right) 0}+b
e^{\left(\frac{3}{2}+\frac{\sqrt{5}}{2}\right) 0}+c
e^{-0}=a+b+c=2
\]
This is the first equation between a, b and c

Answer 10

\[
y'(x)=\left(\frac{3}{2}-\frac{\sqrt{5}}{2}\right) a
e^{\left(\frac{3}{2}-\frac{\sqrt{5}}{2}\right)
x}+\left(\frac{3}{2}+\frac{\sqrt{5}}{2}\right) b
e^{\left(\frac{3}{2}+\frac{\sqrt{5}}{2}\right) x}-c e^{-x}\\
y'(0)=\left(\frac{3}{2}-\frac{\sqrt{5}}{2}\right)
a+\left(\frac{3}{2}+\frac{\sqrt{5}}{2}\right) b-c=2
\]
This is another equation between a, b and c

Answer 11

\[
y''(x)=\left(\frac{3}{2}-\frac{\sqrt{5}}{2}\right)^2 a
e^{\left(\frac{3}{2}-\frac{\sqrt{5}}{2}\right)
x}+\left(\frac{3}{2}+\frac{\sqrt{5}}{2}\right)^2 b
e^{\left(\frac{3}{2}+\frac{\sqrt{5}}{2}\right) x}+c e^{-x}\\

y''(0)=\left(\frac{3}{2}-\frac{\sqrt{5}}{2}\right)^2
a+\left(\frac{3}{2}+\frac{\sqrt{5}}{2}\right)^2 b+c=1
\]
This is the third equation. Is your teacher a masochist?
I refuse to solve these three eqautins by hand for a , b and c

Answer 12

sometimes but shes kind i guest???

Answer 13

I solved them using Mathematica and I got

\[
a= \frac{1}{10} \left(13+5 \sqrt{5}\right)\\b= \frac{13
\sqrt{5}-25}{10 \sqrt{5}}\\c= -\frac{3}{5}
\]

Answer 14

Finally the answer is
\[
y=\frac{1}{10} \left(13+5 \sqrt{5}\right)
e^{\left(\frac{3}{2}-\frac{\sqrt{5}}{2}\right) x}+\frac{\left(13
\sqrt{5}-25\right) e^{\left(\frac{3}{2}+\frac{\sqrt{5}}{2}\right)
x}}{10 \sqrt{5}}-\frac{3 e^{-x}}{5}
\]

Answer 15

I would never assign such a problem to be done by hand.

Answer 16

I have to go. It was nice helping you.

Answer 17

thank you soo much till next time

Answer 18

YW