# Thread: A linear first order equations

1. ## 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.

2. 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:

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

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!!