Linear Differential Equations

Given a 1st order linear differential equation in the normal form

$$y^{\prime} + P(x)y = Q(x)$$

Write in the differential form

$$(P(x)y - Q(x))dx + dy = 0$$

Then

$$M_{y} = P(x), \ N_{x} = 0$$

and $(1/N)[M_{y} - N_{x}] = P(x)$ is a function of $x$ only. Thus,

$$\mu = e^{\int P(x)dx}$$

Now multiplying $\mu(x) = e^{\int P(x)dx}$ to the normal form $y^{\prime} + P(x)y = Q(x)$ to get

$$y^{\prime}e^{\int P(x)dx} + P(x)ye^{\int P(x)dx} = Q(x)e^{\int P(x)dx}$$

The left-hand side is the derivative of $ye^{\int P(x)dx}$. Thus

$$ye^{\int P(x)dx} = \int Q(x)e^{\int P(x)dx}dx + c .$$

Therefore the general solution is given by

$$y = e^{-\int P(x)dx}(\int Q(x)e^{\int P(x)dx}dx + c)$$

The 1st order linear differential equation can be solved by using the above formula. But it is much easier to use the integrating factor $e^{\int P(x)dx}$.

Example 1..16   Find the general solution of $y^{\prime} + \frac{2}{x}y = x$.

SOLUTION We find the integrating factor.

$$\mu = e^{\int \frac{2}{x}}dx = e^{2\log{x}} = x^{2} .$$

Now multiplying the integrating factor to both sides.

$$x^{2}y^{\prime} + 2xy = x^{3} .$$

Now

$$\frac{d(x^{2}y)}{dx} = x^{3} .$$

Integrating both sides to get

$$x^{2}y = \int x^{3}dx = \frac{x^{4}}{4} + c .$$

Thus the general solution is

$$y = \frac{x^{2}}{4} + \frac{c}{x^{2}}\ensuremath{\ \blacksquare}$$

In a RLC circuit, the volatage drop by the circuit current $i$ at the resistance $R$(Ω), at the inductance $L$(H), and at the capacitance $C$(F) is given by

$$1. \ v_{R} = Ri$$

$$2. \ v_{L} = L\frac{di}{dt}$$

$$3. \ v_{C} = \frac{1}{C}\int idt$$

Figure: RLC AC circuit

Then by Kirchhoff'S 2nd law,

$$v_{R} + v_{L} + v_{C} = E .$$

or

$$L\frac{di}{dt} + R_{i} + \frac{1}{C}\int idt = E .$$

Example 1..17   Find the value of corrent $i$ runs through the RL circuit, where $R, L, E$ are constant, $i(0) = I_{0}$

SOLUTION By Kirchhoff's voltage law,

$$L\frac{di}{dt} + Ri = E .$$

Write in the normal form

$$\frac{di}{dt} + \frac{R}{L}i = \frac{E}{L} .$$

Now the integrating factor $\mu$ is $e^{Rt/L}$. Then multiplying $\mu$ to both sides of the equation.

$$e^{Rt/L}(L\frac{di}{dt} + Ri) = \frac{E}{L}e^{Rt/L} .$$

Note that the left-hand side of the equation is the derivative of the integrating factor times the independent variable $i$. Thus

$$d(e^{Rt/L}i) = \frac{E}{L}e^{Rt/L}dt .$$

Thus the general solution is

$$i = \frac{E}{R} + ce^{-Rt/L}$$

Now using the initial value $i(0) = I_{0}$, we have $c = I_{0} - (E/R)$. Therefore,

$$i = \frac{E}{R} + (I_{0} - \frac{E}{R})e^{-Rt/L}\ensuremath{\ \blacksquare}$$