Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (anonymous):

Solve the following DE: Q''(t) + Q'(t)R/L + Q(t)/(LC) = V(t)

OpenStudy (anonymous):

so, as far as I have gotten in my DE class concerning higher order DEs is that for a second order linear DEs with constant coefficients, you can solve for the roots of a quadratic of a similar form (probably a terrible way to word that sorry) so like, for this you would then find the roots of \[r^2 + \frac{R}{L}r + \frac{1}{LC} = 0\]R, L, and C are all constants. but no relation between them is given so I'm really not sure what to do here...

OpenStudy (anonymous):

You can still solve for \(r\). Using the quadratic formula, \[r=\frac{-\frac{R}{L}\pm\sqrt{\frac{R^2}{L^2}-\frac{4}{LC}}}{2}\] Assuming the discriminant is positive, you'll have two exponential solutions. If it's zero, then just one exponential solution. If it's negative, you're dealing with complex roots to the characteristic equation, which will involve two trigonometric/exponential solutions.

OpenStudy (anonymous):

Also, this only takes care of the homogeneous solution. Is \(V(t)=0\) given, or is it something else?

OpenStudy (anonymous):

But, how can I solve it without knowing the values of R, L and C? And not too sure, this is for a different class, I just know that I need to solve the problem using the 2nd-order DE I found.

OpenStudy (anonymous):

If you're not given some explicit domains for those constants, I think you'll have to consider all three cases... We're not going to be able to figure out part of the solution if \(V(t)\) is non-zero.

OpenStudy (anonymous):

And depending on the context of this problem, you could determine whether there are any restrictions on the constants, like \(R>0\).

OpenStudy (anonymous):

Okay, so given that I am fairly certain I am supposed to be finding a general solution for V(t), I'm probably missing something else in this problem... I don't believe R, L, or C can be < 0 since this is the DE for a series RLC circuit (Resistor, Inductor (L), and Capacitor)

OpenStudy (anonymous):

Well, unless you happen to know a lot about signals and systems or circuit analysis, thanks for the help you've offered, but it seems there is something more fundamental that I am missing to be able to solve this problem.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!