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!!!

- anonymous

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- ganeshie8

familiar with the method of undetermined coefficients ?

- anonymous

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

- Michele_Laino

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- ganeshie8

can you use it on below two differential equations separately ?
[(D^2)+D]y = x^2 + 3x
[(D^2)+D]y = e^3x

- anonymous

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

- Michele_Laino

yes!

- ganeshie8

yes only find a particular solution for each of the equations :
[(D^2)+D]y = x^2 + 3x
[(D^2)+D]y = e^3x

- anonymous

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

- Michele_Laino

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

- Michele_Laino

we are applying the superposition principle of solutions

- anonymous

i still can't understand ;;

- Michele_Laino

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

- Michele_Laino

then you have to use the principle of identity of polynomials

- anonymous

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

- ganeshie8

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 !

- Michele_Laino

correct!

- IrishBoy123

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?

- anonymous

how about the B1?

- Michele_Laino

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

- anonymous

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

- Michele_Laino

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

- Michele_Laino

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

- Michele_Laino

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

- anonymous

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

- Michele_Laino

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

- anonymous

ah yes!

- Michele_Laino

ok! correct!

- anonymous

thanks a lot!!!

- Michele_Laino

:)

- anonymous

can i still ask you a question?

- Michele_Laino

yes!

- anonymous

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

- Michele_Laino

is it a series RC circuit?

- anonymous

ohm's law

- Michele_Laino

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

- Michele_Laino

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

- Michele_Laino

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

- Michele_Laino

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

- Michele_Laino

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

- Michele_Laino

those equations can solve your problem

- anonymous

thanks

- Michele_Laino

:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.