# A linear first order equations

• May 19th 2008, 08:29 PM
edeffect
A 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 PM
Chris L T521
Quote:

Originally Posted by edeffect
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.

The differential equation that can be set up as follows:

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

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

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

Let us solve the equivalent homogenous part first:

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

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

Thus, the complementary solution is:

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

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:

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

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

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

Applying the annihilator to both sides, we get:

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

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

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

Thus the solution to the DE is:

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

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

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

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

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

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

$\implies (6B+4A)cos(t)+(-6A+4B)sin(t)=6sin(t)$
$\implies (6B+4A)=0 \ and \ (-6A+4B)=6$

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

Therefore, the solution to the DE becomes:

$Q(t)=c_1e^{-t}+c_2e^{-5t}+\frac{3}{13}\left(-3cos(t)+2sin(t)\right)$
Now apply the initial conditions $Q(0)=.2 \ and \ I(0)=0$.

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

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

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

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

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

Therefore,

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

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