## RolyPoly Obtain a general solution using the method of undetermined coefficients. $y'-3y=xe^{2x}+6$ one year ago one year ago

1. Algebraic!

I love the method of undetermined guessing.

2. RolyPoly

$y_{c}'-3y_c = 0$$y_{c}=ce^{3x}$$\text{Try } y_{p1}=Ce^{2x}$$2Ce^{2x}-3Ce^{2x} = xe^{2x}$$-Ce^{2x} = xe^{2x}$$C = -x$ $y_{p2}=D$$0 - 3D = 6$$D=-2$ $y= ce^{3x}-xe^{2x}-2$?! No!!~~~

3. RolyPoly

''Educated guess'' lol

4. Algebraic!

2Ce2x−3Ce2x=xe2x

5. sirm3d

the particular solution is $\huge y_p=Be^{2x}+Cxe^{2x}+D$

6. Algebraic!

going to need something like e^2x (A+Bx)

7. RolyPoly

Hmm... Let me try again!!!

8. RolyPoly

$y_{p1} = e^{2x}(A+Bx)$$2e^{2x}(A+Bx)+Be^{2x}-3e^{2x}(A+Bx)=xe^{2x}$$Be^{2x}-e^{2x}(A+Bx)=xe^{2x}$$e^{2x}(A-B+Bx)=-xe^{2x}$$A=B=-1$ $y_{p1} = e^{2x}(-1-x)$ $y_{p2} = -2$ $y= ce^{3x} -e^{2x}-xe^{2x}-2$ It works!!! o_O

9. sirm3d

nice work.

10. RolyPoly

How did you get this: $$y_p=Be^{2x}+Cxe^{2x}+D$$ ??

11. sirm3d

if the right hand side is $\large e^{2x}$ the corresponding particular solution is $\large Ce^{2x}$ but if the RHS is some expression $\large x^{n}e^{2x}$ then the particular solution is $\huge \left( C_0 + C_1x + C_2 x^2 + ... + C_n x^n\right) e^{2x}$

12. RolyPoly

Does the coefficient of x in e^{nx} matter?

13. sirm3d

yes. that determines the particular solution

14. RolyPoly

Hmm... In what way?

15. sirm3d

the polynomial factor of e^{ax}, that is $\large c_0 + c_1x + c_2x^2 + ... c_nx^n$

16. RolyPoly

Why is particular solution for this one is e^(2x) (A+Bx) But it doesn't work for y' - 3y = xe^(3x) + 4???

17. sirm3d

it will not work since the complementary solution is $\large c_1e^{3x}$. the constant coefficient is taken by the complementary solution, therefore the polynomial factor of the particular solution "begins with" $\large Bx$. Thus the particular solution is $\large (Bx + Cx^2)e^{3x}$

18. RolyPoly

That's complicated :(