Second-order differential equations (electrical circuit)

[EngStudentNcl]

Homework Statement

The current I(t) in an electrical circuit satisfies the second-order differential equation 2Ϊ + 3İ + I = 0.
Find the solution I(t) of this equation which also satisfies the initial conditions I(0) = 1, İ(0) = 0.

Homework Equations

a((d^2y)/(dx^2)) + b(dy/dx) + cy = 0
Where a, b and c are constant coefficients. It can be shown that the solution to this equation is y = (Ae^m)(1^x) + (Be^m)(2^x)

Where A and B are two arbitrary constants and m1 and m2 are the roots of the quadratic
equation am2 + bm + c = 0
This quadratic equation is called the auxiliary equation.

The Attempt at a Solution

Really stuck on this one. Presumably the equation is 2((d^2y)/(dx^2)) + 3(dy/dx) + 1 = 0, giving the auxillary equation 2m^2 + 3m + 1 = 0.

It's here where I'm stuck (though I'm not even sure the above is correct). We've been taught differentiation, but not how to apply it! I've spent too much time on this already and I have to get revision done for my exams. The question goes on to say...

Solve the auxiliary equation

Therefore m =

So I(t) =

Given initial condition I(0) = 1, therefore

Differentiate to give İ(t) =

Given initial condition İ(0) = 0


Solution is I(t) = 2e^-0.5t – e^-t

Any help massively appreciated.

 

[micromass]

Hello EngStudentNcl!

It seems that first you need to solve the equation

2m^2 + 3m + 1 = 0

this has nothing to do with differentiation, you simply need to apply the quadratic formula.

 

[ideasrule]

EDIT: oops, edit conflict with micromass

I'm still not sure where you're stuck. Your auxiliary equation is correct. Do you know how to solve it? There are no tricks involved--it's just a quadratic.

Once you've got that, the general solution is I = (Ae^m1*t) + (Be^m2*t), where m1 and m2 are the two solutions you got. Now you use the 2 boundary conditions to find the 2 unknowns, A and B, and that's the answer.

As a side note, there's nothing preventing m1 and m2 from being complex; they just happen to be real for this problem.

 

[EngStudentNcl]

Ah, thank you both! I've been looking at it, on the paper there's quite a gap between 'Given initial condition İ(0) = 0' and 'Solution is...', I was (obviously mistakenly) presuming there was a further (mathematical) step between solving the quadratic & the actual answer.

My apologies, I've been making it a lot more complex than it actually is. Exam stress I guess.

Thanks again, that's a massive weight off my mind (and another lesson in not overcomplicating things)!