# Thread: First Order Linear Diff. Eqn?

1. ## First Order Linear Diff. Eqn?

Hi all,

I am unsure of what to do with this one.

I have been given some conditions in an electric circuit. Due to Kirchoff's law I have calculated the differential equation to be:

2(dI/dt) + 12(dQ/dt) + 10Q = 12 sin(t)

where 12 sin(t) is the supplied voltage.

Im supposed to find Q at time t with initial conditions Q(0) = 0.2 and I(0)=0.

Is this a first order linear differential equation? Or is it classified as something else?

What confuses me is that it has 2(dI/dt) + 12(dQ/dt). I was thinking that perhaps dI/dt = d^2Q/dt^2 or vise versa but really im not sure.

2. Actually yes,

dI/dt = d^2Q/dt^2

so the equation is

2(d^2Q/dt^2) + 12(dQ/dt) + 10Q = 12 sin(t)

which is a second order linear differential equation.

3. [quote=woody198403;148103]Actually yes,

dI/dt = d^2Q/dt^2

so the equation is

2(d^2Q/dt^2) + 12(dQ/dt) + 10Q = 12 sin(t)

This is a 2nd order linear DE

So first we need to solve the associated homogenious equation

$2Q''+12Q'+10Q=0$

$2m^2+12m+10=0 \iff 2(m+1)(m+5)=0$

so the two roots are m=-1 and m=-5

This gives the complimentry solution $y=c_1e^{-t}+c_2e^{-5t}$

Now we need to find the particular solution

since the driving function is 12sin(t)

we know the particular solution is of the form

$y=A\sin(t)+B\cos(t)$

taking a few derivatives and plugging into the non homogenious equation and solvinf for A and B will finish the problem.

Good luck.

4. Originally Posted by woody198403
Hi all,

I am unsure of what to do with this one.

I have been given some conditions in an electric circuit. Due to Kirchoff's law I have calculated the differential equation to be:

2(dI/dt) + 12(dQ/dt) + 10Q = 12 sin(t)

where 12 sin(t) is the supplied voltage.

Im supposed to find Q at time t with initial conditions Q(0) = 0.2 and I(0)=0.

Is this a first order linear differential equation? Or is it classified as something else?

What confuses me is that it has 2(dI/dt) + 12(dQ/dt). I was thinking that perhaps dI/dt = d^2Q/dt^2 or vise versa but really im not sure.