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anonymous
 one year ago
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
 one year ago
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!!!

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2familiar with the method of undetermined coefficients ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep. but i'm quite confused on how to use that

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3for the first one, try to search a solution like this: \[A{e^{3x}} + {B_1} + {B_2}x + {B_3}{x^2} + {B_4}{x^3} + {B_5}{x^4}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2can you use it on below two differential equations separately ? [(D^2)+D]y = x^2 + 3x [(D^2)+D]y = e^3x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is this right? [(D^2)+D]y = e^3x y= c1+c2e^x + (1/12)e^3x

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2yes only find a particular solution for each of the equations : [(D^2)+D]y = x^2 + 3x [(D^2)+D]y = e^3x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but i don't know how to solve for this [(D^2)+D]y = x^2 + 3x

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3you 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
 one year ago
Best ResponseYou've already chosen the best response.3we are applying the superposition principle of solutions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i still can't understand ;;

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3you 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
 one year ago
Best ResponseYou've already chosen the best response.3then you have to use the principle of identity of polynomials

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is this right? B2=1 B3=1/2 B4= 1/3 B5=0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2you 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 !

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0the first one can be made less boring by multiplying by \(e^x\) and seeing the LHS as \( (e^x y^{\prime})^{\prime} \), n'estce pas?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3\(B_1\) is added to \(c_1\) so we get the new constant: \(k_1=c_1+B_1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks to all!!! can you still help me for the second one y'' + 4y = 8cos2x 4x??

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3for 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
 one year ago
Best ResponseYou've already chosen the best response.3or, 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
 one year ago
Best ResponseYou've already chosen the best response.3please 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
 one year ago
Best ResponseYou've already chosen the best response.0is this the answer to the second one? y=c1cos2x+c2sin2x+2xsinxx

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3do you mean: y=c1cos2x+c2sin2x+2xsin(2x)x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can i still ask you a question?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks. 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
 one year ago
Best ResponseYou've already chosen the best response.3is it a series RC circuit?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3we can write the subsequent ODE: \[  \frac{Q}{C} + RI = {E_0}\cos \left( {\omega t} \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3or, more precisely, we have this: \[\frac{{dV}}{{dt}} + \frac{V}{{RC}} = 0\] where \(V \) is the electric voltage

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3\(V= V(t)\) is the unknown function, of course

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3furthermore we have this equation: \[I =  C\frac{{dV}}{{dt}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3and, of course: \[Q\left( t \right) = CV\left( t \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3those equations can solve your problem
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