Question

some easy auxilary...

Answer 1

\(\large\color{#000000 }{ \displaystyle y''-4y'+3y=x^2 }\)

Answer 2

So, at first, my roots for the auxilary \(\large\color{#000000 }{ \displaystyle r^2-4r+3=0 }\)
are: \(\large\color{#000000 }{ \displaystyle r=1,3 }\)

Answer 3

Then, the function G(x) is a quadratic polynomial, so y\(_p\)(x) is
\(\large\color{#000000 }{ \displaystyle y_p(x)=ax^2+bx+c }\)

And then we find the derivatives accordingly...

Answer 4

Yep, now it's just number crunching.

Answer 5

yes, number crunching is the essence of the experience in all maths:)

Answer 6

\(\large\color{#000000 }{ \displaystyle y_p~'(x)=2ax+b }\)
\(\large\color{#000000 }{ \displaystyle y_p~''(x)=2a }\)

Answer 7

\(\large\color{#000000 }{ \displaystyle 2a+2ax+b-2(ax^2+bx+c)=x^2 }\)
\(\large\color{#000000 }{ \displaystyle -2ax^2+(2a-2b)x+(2a+b-2c)=x^2 }\)

\(\large\color{#000000 }{ \displaystyle -2a=1 }\)
\(\large\color{#000000 }{ \displaystyle 2a-2b=0 }\)
\(\large\color{#000000 }{ \displaystyle 2a+b-2c=0 }\)

Answer 8

comapring coefficients, ...

Answer 9

Careful, you made a mistake. It's -4y'

Answer 10

And +3y(x). Your working suggests the question is y''(x)+y'(x)-2y(x) when it's in fact y''(x)-4y'(x)+3y(x)

Answer 11

So it becomes \[2x-4(2ax+b)+3(ax^2+bx+c)=x^2\]

Answer 12

yeah, true that ...

Answer 13

I mixed the coefficients,

\(\large\color{#000000 }{ \displaystyle 2x-4(2ax+b)+3(ax^2+bx+c)=x^2 }\)
\(\large\color{#000000 }{ \displaystyle 2x-8ax-4b+3ax^2+3bx+3c=x^2 }\)
\(\large\color{#000000 }{ \displaystyle 3ax^2+2x-8ax+3bx-4b+3c=x^2 }\)
\(\large\color{#000000 }{ \displaystyle 3ax^2+x(2-8a+3b)+(3c-4b)=x^2 }\)

Answer 14

0a-4b+3c=0
2-8a+3b=0
3a=1

Answer 15

was thinking to make it a matrix that is why I wrote it like this....
will simply solve it.

Answer 16

I've gotta go now, but if you want to check your answer when you finish then here's what I think the solution is:

https://latex.codecogs.com/gif.latex?y%28x%29%3DAe%5Ex+Be%5E%7B3x%7D+%5Cfrac%7B1%7D%7B3%7Dx%5E2-%5Cfrac%7B8%7D%7B9%7Dx-%5Cfrac%7B38%7D%7B27%7D

Didn't want to write it here and spoil it for you.

Answer 17

a=1/3

2-8(1/3)+3b=0
6/3-8/3=-3b
2/9=b

-4(2/9)+3c=0
8/27=c


a=1/3
2/9=b
8/27=c

Answer 18

\(\color{#000000 }{ \displaystyle y_c(x)=c_1e^{x}+c_2e^{3x} }\)

\(\color{#000000 }{ \displaystyle y_p(x)=(1/3)x^2+(2/9)x+(8/27) }\)

So,
\(\color{#000000 }{ \displaystyle y=y_c+ y_p=c_1e^{x}+c_2e^{3x} +(1/3)x^2+(2/9)x+(8/27) }\)

Answer 19

our coefficients differ

Answer 20

The equations you're working off are wrong. You've said y''(x)=2x when it's 2a

Answer 21

yes, another errr. tnx

Answer 22

I will rework it.

Answer 23

I'm really going this time, lol. Good luck with the rest of the question, I'll be back in the morning.

Answer 24

\(\color{#000000 }{ \displaystyle 2a-4(2ax+b)+3(ax^2+bx+c)=x^2 }\)
\(\color{#000000 }{ \displaystyle 3ax^2-8ax+3bx+3c+2a-4b=x^2 }\)

Answer 25

\(\color{#000000 }{ \displaystyle 3a=1,~\Longrightarrow ~a=1/3 }\)
\(\color{#000000 }{ \displaystyle -8a+3b=0,~\Longrightarrow ~3b=8(1/3),~\Longrightarrow ~b=8/9 }\)

\(\color{#000000 }{ \displaystyle 3c+2a-4b=0,~\Longrightarrow 3c+2(1/3)-4(8/9)=0 \\[0.5em] 3c+2/3-32/9=0,~\Longrightarrow c=26/27}\)

Answer 26

\(\Large{\bbox[4px, #ffcc99 ,border:2px solid black ]{ \color{#800000}{\displaystyle y=c_1e^{x}+c_2e^{3x} +\frac{1}{3}x^2+\frac{8}{9}x+\frac{26}{27}} }}\)

Answer 27

Ah yeah I wingspaned up, missed a minus sign on my b and it ruined my c. Nice work.