anonymous
  • anonymous
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!!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ganeshie8
  • ganeshie8
familiar with the method of undetermined coefficients ?
anonymous
  • anonymous
yep. but i'm quite confused on how to use that
Michele_Laino
  • Michele_Laino
for the first one, try to search a solution like this: \[A{e^{3x}} + {B_1} + {B_2}x + {B_3}{x^2} + {B_4}{x^3} + {B_5}{x^4}\]

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More answers

ganeshie8
  • 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
  • anonymous
is this right? [(D^2)+D]y = e^3x y= c1+c2e^-x + (1/12)e^3x
Michele_Laino
  • Michele_Laino
yes!
ganeshie8
  • 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
  • anonymous
but i don't know how to solve for this [(D^2)+D]y = x^2 + 3x
Michele_Laino
  • 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
  • Michele_Laino
we are applying the superposition principle of solutions
anonymous
  • anonymous
i still can't understand ;;
Michele_Laino
  • 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
  • Michele_Laino
then you have to use the principle of identity of polynomials
anonymous
  • anonymous
Is this right? B2=-1 B3=1/2 B4= 1/3 B5=0
ganeshie8
  • 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
  • Michele_Laino
correct!
IrishBoy123
  • 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
  • anonymous
how about the B1?
Michele_Laino
  • Michele_Laino
\(B_1\) is added to \(c_1\) so we get the new constant: \(k_1=c_1+B_1\)
anonymous
  • anonymous
thanks to all!!! can you still help me for the second one y'' + 4y = 8cos2x -4x??
Michele_Laino
  • 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
  • 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
  • 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
  • anonymous
is this the answer to the second one? y=c1cos2x+c2sin2x+2xsinx-x
Michele_Laino
  • Michele_Laino
do you mean: y=c1cos2x+c2sin2x+2xsin(2x)-x
anonymous
  • anonymous
ah yes!
Michele_Laino
  • Michele_Laino
ok! correct!
anonymous
  • anonymous
thanks a lot!!!
Michele_Laino
  • Michele_Laino
:)
anonymous
  • anonymous
can i still ask you a question?
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • 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
  • Michele_Laino
is it a series RC circuit?
anonymous
  • anonymous
ohm's law
Michele_Laino
  • Michele_Laino
we can write the subsequent ODE: \[ - \frac{Q}{C} + RI = {E_0}\cos \left( {\omega t} \right)\]
Michele_Laino
  • Michele_Laino
or, more precisely, we have this: \[\frac{{dV}}{{dt}} + \frac{V}{{RC}} = 0\] where \(V \) is the electric voltage
Michele_Laino
  • Michele_Laino
\(V= V(t)\) is the unknown function, of course
Michele_Laino
  • Michele_Laino
furthermore we have this equation: \[I = - C\frac{{dV}}{{dt}}\]
Michele_Laino
  • Michele_Laino
and, of course: \[Q\left( t \right) = CV\left( t \right)\]
Michele_Laino
  • Michele_Laino
those equations can solve your problem
anonymous
  • anonymous
thanks
Michele_Laino
  • Michele_Laino
:)

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