Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

Obtain a general solution using the method of undetermined coefficients. \[y'-3y=xe^{2x}+6\]

Differential Equations
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

I love the method of undetermined guessing.
\[y_{c}'-3y_c = 0\]\[y_{c}=ce^{3x}\]\[\text{Try } y_{p1}=Ce^{2x}\]\[2Ce^{2x}-3Ce^{2x} = xe^{2x}\]\[-Ce^{2x} = xe^{2x}\]\[C = -x\] \[y_{p2}=D\]\[0 - 3D = 6\]\[D=-2\] \[y= ce^{3x}-xe^{2x}-2\]?! No!!~~~
''Educated guess'' lol

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

2Ce2x−3Ce2x=xe2x
the particular solution is \[\huge y_p=Be^{2x}+Cxe^{2x}+D\]
going to need something like e^2x (A+Bx)
Hmm... Let me try again!!!
\[y_{p1} = e^{2x}(A+Bx)\]\[2e^{2x}(A+Bx)+Be^{2x}-3e^{2x}(A+Bx)=xe^{2x}\]\[Be^{2x}-e^{2x}(A+Bx)=xe^{2x}\]\[e^{2x}(A-B+Bx)=-xe^{2x}\]\[A=B=-1\] \[y_{p1} = e^{2x}(-1-x)\] \[y_{p2} = -2\] \[y= ce^{3x} -e^{2x}-xe^{2x}-2\] It works!!! o_O
nice work.
How did you get this: \(y_p=Be^{2x}+Cxe^{2x}+D\) ??
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 \[\huge \left( C_0 + C_1x + C_2 x^2 + ... + C_n x^n\right) e^{2x}\]
Does the coefficient of x in e^{nx} matter?
yes. that determines the particular solution
Hmm... In what way?
the polynomial factor of e^{ax}, that is \[\large c_0 + c_1x + c_2x^2 + ... c_nx^n\]
Why is particular solution for this one is e^(2x) (A+Bx) But it doesn't work for y' - 3y = xe^(3x) + 4???
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 \[\large (Bx + Cx^2)e^{3x}\]
That's complicated :(

Not the answer you are looking for?

Search for more explanations.

Ask your own question