Question

Find a particular solution of y"-8y'+16y=23cosx-7sinx.

Answer 1

@amistre64 @freckles @mathstudent55 @ParthKohli @uri

Answer 2

what is your homogenous solution?

Answer 3

The answer is \[y _{p}=\cos x-\sin x\]. But how do I do this problem?

Answer 4

.... how i do it is ... i start by finding the =0 solution, the homogenous solution

Answer 5

r=4, double root.
y=e^(4x)(c1+c2x)

Answer 6

now work a variation of parameters, let c1 and c2 be functions of x, and plug it into the equation

Answer 7

I need to use a method involving the substitution of yp.

Answer 8

let

\[y_t=e^{4x}(f(x)+xg(x))\]

what is the 1st and second derivatives?

Answer 9

What's that ^ method above? Because I wanted to use the substitution \[y _{p}=Acos x-Bsin x\]

Answer 10

variation of parameters ...

Answer 11

your method looks like trial and error ... the guessing method

Answer 12

16f e^(4x) + 16gx e^(4x)

-32f e^(4x) -8ge^(4x) -32gx e^(4x) +(f'e^(4x)+ g'x e^(4x))
( = 0 )

4f' e^(4x) + 16f e^(4x) + g'e^(4x) + 4ge^(4x) + 4ge^(4x)+ 16gx e^(4x)+ 4g'x e^(4x)

----------------------------

32f e^(4x) + 32gx e^(4x)

-32f e^(4x) -8ge^(4x) -32gx e^(4x) +(f'e^(4x)+ g'x e^(4x))
( = 0 )

4f' e^(4x) + g'e^(4x) + 8ge^(4x) + 4g'x e^(4x)
---------------------------


f'e^(4x)+ g'x e^(4x) = 0


f' 4e^(4x) + g'(e^(4x) + 4x e^(4x)) = 23cos(x)-7sin(x)

Answer 13

@Idealist10 check the solution

Answer 14

f'+ g'x = 0

4f' + g'(1+ 4x) = 23cos(x)-7sin(x)

Answer 15

@Idealist10 did you check the attachment?

Answer 16

Yes, I did. Thanks a lot for the help, guys.

Answer 17

since f' = -g'x

-4g'x + g'+ 4g'x = 23cos(x)-7sin(x)

f' = -23x cos(x) + 7x sin(x)
g' = 23cos(x) - 7sin(x)

gx = 23xsin(x) + 7xcos(x)
f = -23xsin(x) +7 sin(x)-7 x cos(x)-23 cos(x)

f+gx = 7sin(x)-23cos(x)

Answer 18

pfft, mine got off track someplace in there :)

Answer 19

@amistre64 did you check my method?

Answer 20

im not familiar enough with your method to comment critically about it

Answer 21

im barely familiar with my own thougths :)

Answer 22

Its shortcut methods for variation of parameters

Answer 23

ive seen it, but i just do not have the practical experience with it

Answer 24

Hmm we use it when we have sine or cosine functions as the tranformation to e^ix is easy//
for example 1/f(D) e^ax is 1/f(a) e^ax if f(a) is not zero