Differential equations

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 our expert's

answer on brainly

SEE EXPERT ANSWER

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

A community for students.

Differential equations

Mathematics
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

give question......
wt of them??
\[y"+3y'-4y = 0\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

if we let \[y= e^{mx}\]
I know that we can get \[(m^2 +3m-4)e^{mx}=0\]
use characteristic equation: m^2+3m-4=0 m=-4 and m=1 so y=c1e^(-4t)+c2e^t
(d^2+3d-4)y=0 d=4,-1 y=ae^4x+be^-x
and it factors into \[(m-1)(m+4)=0\]
so since the roots are real and distinct, y = \[ae^{x}+be^{-4x}\]
CORRECT
it will be ae^4x+be^-x.
but what if \[y"-10y'+25y = 0 \]
because your char. equation m^2+3m-4=0 has the roots 4,-1.
(a+bx)e^5x
Then you have two same roots m=5 and then y=c1te^5t+c2e^5t
how do we proceed the problem? do we let \[y= xe^{mx}\]
"because your char. equation m^2+3m-4=0 has the roots 4,-1" dipank this is NOT CORRECT
yes i checked........sorry for that........
noproblem, i just didn't want to confuse..
for the other root I think there was another way to solve it when the roots are equal and real, the formula was different and I think dipank sounds right but I am not sure how to prove it
long proof..it'd be in your book for sure. circuit theory is the only way i ever wrapped my brain around this concept..a critically damped RLC circuit is modelled by a characteristic equation with two real same roots. Good luck with your proof and further studies though.
here is a link that might help you understand repeated roots in characteristic equation http://tutorial.math.lamar.edu/Classes/DE/RepeatedRoots.aspx

Not the answer you are looking for?

Search for more explanations.

Ask your own question