Hello I'm trying to solve the following mathematical problem, but keep getting stuck. If someone could solve the question and show me their working, I'm sure I would learn a lot.

Thanks.

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- May 19th 2008, 08:29 PMedeffectA linear first order equations
Hello I'm trying to solve the following mathematical problem, but keep getting stuck. If someone could solve the question and show me their working, I'm sure I would learn a lot.

Thanks. - May 19th 2008, 09:11 PMChris L T521
The differential equation that can be set up as follows:

$\displaystyle L\frac{dI}{dt}+R\frac{dQ}{dt}+\frac{1}{C}Q=E$

Given the values for R, L, C, and E, and noting that $\displaystyle \frac{dI}{dt}=\frac{d^2Q}{dt^2}$, we get the DE:

$\displaystyle \frac{d^2Q}{dt^2}+6\frac{dQ}{dt}+5Q=6sin(t)$

Let us solve the equivalent homogenous part first:

Assuming a solution of $\displaystyle Q=e^{rt}$, we have the following characteristic equation:

$\displaystyle r^2+6r+5=0 \implies r=-5 \ or \ r=-1$.

Thus, the complementary solution is:

$\displaystyle Q_c=c_1e^{-t}+c_2e^{-5t}$.

Now solve the non-homogeneous equation.

The technique I will use is known as the annihilator approach. I wasn't taught another way to solve these equations (besides Variation of Parameters, but I believe that it is unnecessary here).

First, we need to get the DE in Differential Operator Notation:

$\displaystyle \left(D^2+6D+5\right)(Q)=6sin(t)$

I will leave it for you (and others) to show that $\displaystyle (D^2-2\alpha D+(\alpha^2+\beta^2))^n$ annihilates $\displaystyle x^{n-1}e^{\alpha x}sin(\beta x)$

Therefore, $\displaystyle D^2+1$ annihilates $\displaystyle 6sin(t)$.

Applying the annihilator to both sides, we get:

$\displaystyle (D^2+1)(D^2+6D+5)(Q)=0$

Convert the equation so it has the form of the characteristic equation:

$\displaystyle (r^2+1)(r^2+6r+5)=0 \implies r=\pm i, \ r=-1 \ or \ r=-5$

Thus the solution to the DE is:

$\displaystyle Q=c_1e^{-t}+c_2e^{-5t}+c_3cos(t)+c_4sin(t)$

where $\displaystyle c_3cos(t)+c_4sin(t)$ is the particular solution. The particular solution can't have arbitrary constants. To find $\displaystyle c_3$ and $\displaystyle c_4$, which I will denote by $\displaystyle A$ and $\displaystyle B$ respectively, plug $\displaystyle Q_p$ into the original DE.

$\displaystyle Q_p=Acos(t)+Bsin(t)$

$\displaystyle Q_{p}^{/}=-Asin(t)+Bcos(t)$

$\displaystyle Q_{p}^{//}=-Acos(t)-Bsin(t)$

$\displaystyle \therefore (-Acos(t)-Bsin(t))+6(-Asin(t)+Bcos(t))+5(Acos(t)+Bsin(t))=6sin(t)$

$\displaystyle \implies (6B+4A)cos(t)+(-6A+4B)sin(t)=6sin(t)$

$\displaystyle \implies (6B+4A)=0 \ and \ (-6A+4B)=6$

You will find that $\displaystyle A=-\frac{9}{13} \ and \ B=\frac{6}{13}$

Therefore, the solution to the DE becomes:

$\displaystyle Q(t)=c_1e^{-t}+c_2e^{-5t}+\frac{3}{13}\left(-3cos(t)+2sin(t)\right)$

Now apply the initial conditions $\displaystyle Q(0)=.2 \ and \ I(0)=0$.

$\displaystyle .2=c_1+c_2-\frac{9}{13}$

Find $\displaystyle I=\frac{dQ}{dt}$:

$\displaystyle Q^{/}(t)=-c_1e^{-t}-5c_2e^{-5t}+\frac{3}{13}\left(3sin(t)+2cos(t)\right)$

$\displaystyle 0=-c_1-5c_2+\frac{6}{13}$.

$\displaystyle \implies c_1=1 \ and \ c_2=-\frac{7}{65}$

Therefore,

$\displaystyle \color{red}\boxed{Q(t)=e^{-t}-\frac{7}{65}e^{-5t}+\frac{3}{13}\left(3cos(t)+2sin(t)\right)}$ coulombs

Woah...that was long. I hope you can absorb all of this!! :D