Question

I need to find the solution for these differential equations:

1) [(D^2)+D]y = x^2 + 3x + e^3x
2) y'' + 4y = 8cos2x -4x

I know their complementary functions which are
1) yc = c1+c2e^-x
2) yc = c1cos2x +c2sin2x

But I don't know how to get the particular functions to find the solution. Please help!!!

Answer 1

familiar with the method of undetermined coefficients ?

Answer 2

yep. but i'm quite confused on how to use that

Answer 3

for the first one, try to search a solution like this:
\[A{e^{3x}} + {B_1} + {B_2}x + {B_3}{x^2} + {B_4}{x^3} + {B_5}{x^4}\]

Answer 4

can you use it on below two differential equations separately ?

[(D^2)+D]y = x^2 + 3x

[(D^2)+D]y = e^3x

Answer 5

is this right?


[(D^2)+D]y = e^3x
y= c1+c2e^-x + (1/12)e^3x

Answer 6

yes!

Answer 7

yes only find a particular solution for each of the equations :

[(D^2)+D]y = x^2 + 3x

[(D^2)+D]y = e^3x

Answer 8

but i don't know how to solve for this [(D^2)+D]y = x^2 + 3x

Answer 9

you have to consider a solution like this one:
\[{B_1} + {B_2}x + {B_3}{x^2} + {B_4}{x^3} + {B_5}{x^4}\]

Answer 10

we are applying the superposition principle of solutions

Answer 11

i still can't understand ;;

Answer 12

you have to replace \(y'\) and \(y''\) with the first and second derivative of \({B_1} + {B_2}x + {B_3}{x^2} + {B_4}{x^3} + {B_5}{x^4}\)

Answer 13

then you have to use the principle of identity of polynomials

Answer 14

Is this right?
B2=-1
B3=1/2
B4= 1/3
B5=0

Answer 15

you may use wolfram to double check
http://www.wolframalpha.com/input/?i=solve+y%27%27%2By%27+%3D+x%5E2+%2B+3x+%2B+e%5E%283x%29

it seems you have got it correctly! good job !

Answer 16

correct!

Answer 17

the first one can be made less boring by multiplying by \(e^x\) and seeing the LHS as \( (e^x y^{\prime})^{\prime} \), n'est-ce pas?

Answer 18

how about the B1?

Answer 19

\(B_1\) is added to \(c_1\) so we get the new constant: \(k_1=c_1+B_1\)

Answer 20

thanks to all!!! can you still help me for the second one y'' + 4y = 8cos2x -4x??

Answer 21

for the second one, you have to try a solution like this one:
\[{A_1}\cos \left( {2x} \right) + {A_2}\sin \left( {2x} \right) + {B_1} + {B_2}x + {B_3}{x^2} + {B_4}{x^3}\]

Answer 22

or, if you prefer, you can split the problem, as @ganeshie8 has indicated before, and then try these solutions separately:
\[{A_1}\cos \left( {2x} \right) + {A_2}\sin \left( {2x} \right)\]
\[{B_1} + {B_2}x + {B_3}{x^2} + {B_4}{x^3}\]

Answer 23

please wait, I have made a typo, the right solution is:
\[x\left\{ {{A_1}\cos \left( {2x} \right) + {A_2}\sin \left( {2x} \right)} \right\} + {B_1} + {B_2}x + {B_3}{x^2} + {B_4}{x^3}\]

Answer 24

is this the answer to the second one?
y=c1cos2x+c2sin2x+2xsinx-x

Answer 25

do you mean:
y=c1cos2x+c2sin2x+2xsin(2x)-x

Answer 26

ah yes!

Answer 27

ok! correct!

Answer 28

thanks a lot!!!

Answer 29

:)

Answer 30

can i still ask you a question?

Answer 31

yes!

Answer 32

Thanks. This is still a problem for differential equations. I have trouble solving these type of problems.

Find the constant EMF E if C=(1/20)farad, R=20 ohms, q=0 when t=0, and q=2 coulomb when t=1 sec

Answer 33

is it a series RC circuit?

Answer 34

ohm's law

Answer 35

we can write the subsequent ODE:
\[ - \frac{Q}{C} + RI = {E_0}\cos \left( {\omega t} \right)\]

Answer 36

or, more precisely, we have this:
\[\frac{{dV}}{{dt}} + \frac{V}{{RC}} = 0\]
where \(V \) is the electric voltage

Answer 37

\(V= V(t)\) is the unknown function, of course

Answer 38

furthermore we have this equation:
\[I = - C\frac{{dV}}{{dt}}\]

Answer 39

and, of course:
\[Q\left( t \right) = CV\left( t \right)\]

Answer 40

those equations can solve your problem

Answer 41

thanks

Answer 42

:)