## 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!!!

1. ganeshie8

familiar with the method of undetermined coefficients ?

2. anonymous

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

3. Michele_Laino

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

4. ganeshie8

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

5. anonymous

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

6. Michele_Laino

yes!

7. 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

8. anonymous

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

9. Michele_Laino

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

10. Michele_Laino

we are applying the superposition principle of solutions

11. anonymous

i still can't understand ;;

12. Michele_Laino

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

13. Michele_Laino

then you have to use the principle of identity of polynomials

14. anonymous

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

15. 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 !

16. Michele_Laino

correct!

17. 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?

18. anonymous

how about the B1?

19. Michele_Laino

$$B_1$$ is added to $$c_1$$ so we get the new constant: $$k_1=c_1+B_1$$

20. anonymous

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

21. Michele_Laino

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

22. Michele_Laino

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

23. Michele_Laino

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

24. anonymous

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

25. Michele_Laino

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

26. anonymous

ah yes!

27. Michele_Laino

ok! correct!

28. anonymous

thanks a lot!!!

29. Michele_Laino

:)

30. anonymous

can i still ask you a question?

31. Michele_Laino

yes!

32. 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

33. Michele_Laino

is it a series RC circuit?

34. anonymous

ohm's law

35. Michele_Laino

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

36. Michele_Laino

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

37. Michele_Laino

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

38. Michele_Laino

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

39. Michele_Laino

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

40. Michele_Laino

those equations can solve your problem

41. anonymous

thanks

42. Michele_Laino

:)