Lesson 6 - Linear First-Order Equations

The standard form of linear first order

d y + P (x) y = Q (x)

When solving for a differential equation of this form, the general solution is as follows:

y (x) = \frac{1}{μ (x)} [\int μ (x) Q (x) d x + C]

where, $μ (x) = e^{\int P (x) d x}$ , which is also known as our integrating factor.

For existence and uniqueness of a linear first-order equation, we analyze the functions of $x$ . If $P (x)$ and $Q (x)$ are continuous on an interval $(a, b)$ that contains the point $x_{0}$ , then the IVP will have a unique solution $y (x)$ in the interval $(a, b)$ .

Example

x y^{'} + 2 y = x^{2} - x - 1, y (1) = \frac{1}{2}

The first thing we must do here is analyze whether the DE is in the standard form. We can see here that the leading term contains an $x$ . We can divide the entire equation by x to ensure proper form.

y^{'} + \frac{2}{x} y = x - 1 - \frac{1}{x}

Notice that I chose to keep $P (x)$ in the form $\frac{2}{x}$ . This will help us later. On the right, if we split up the fraction $\frac{x^{2} - x - 1}{x}$ up into three parts, we end up with $\frac{x^{2}}{x} - \frac{x}{x} - \frac{1}{x}$ . We end up with the simplification above.

From here, we can identify that $P (x) = \frac{2}{x}$ . We will use this to find $μ (x)$ .

μ (x) = e^{\int \frac{2}{x} d x}

\int \frac{2}{x} d x = 2 \int \frac{1}{x} d x = 2 \ln | x |

μ (x) = e^{2 \ln | x |}

From our properties of $\ln$ , we know that $b \ln | a | = \ln | a^{b} |$ . We also know that $e^{\ln | a |} = a$ . Using these properties, we find that

e^{2 \ln | x |} = e^{\ln | x^{2} |} = x^{2}

Now, we can find our general solution $y (x)$ . Recall $y (x)$ :

y (x) = \frac{1}{μ (x)} [\int μ (x) Q (x) d x + C]

Also recall $Q (x) = x - 1 - \frac{1}{x}$

y (x) = \frac{1}{x^{2}} [\int x^{2} (x - 1 - \frac{1}{x}) d x + C]

Let's solely focus on the integral, which can be simplified.

\int x^{2} (x - 1 - \frac{1}{x} d x) = \int (x^{3} - x^{2} - x) d x

= \frac{1}{4} x^{4} - \frac{1}{3} x^{3} - \frac{1}{2} x^{2}

Now, let's continue by multiple the integrating factor by the fully integrated equation and our constant $C$ .

\frac{1}{x^{2}} (\frac{1}{4} x^{4} - \frac{1}{3} x^{3} - \frac{1}{2} x^{2} + C) = \frac{1}{4} x^{2} - \frac{1}{3} x - \frac{1}{2} + \frac{C}{x^{2}}

y (x) = \frac{1}{4} x^{2} - \frac{1}{3} x - \frac{1}{2} + \frac{C}{x^{2}}

Now, simply plug in our initial conditions to find $C$ .

\frac{1}{2} = \frac{1}{4} (1)^{2} - \frac{1}{3} (1) - \frac{1}{2} + \frac{C}{(1)^{2}}

\frac{1}{2} = \frac{1}{4} - \frac{1}{3} - \frac{1}{2} + C

\frac{1}{2} - \frac{1}{4} + \frac{1}{3} + \frac{1}{2} = C

C = \frac{13}{12}

Our final solution is:

y (x) = \frac{1}{4} x^{2} - \frac{1}{3} x - \frac{1}{2} + \frac{13}{12} x^{- 1}

Try for Yourself:

$t^{2} \frac{d x}{d t} + 3 t x = t^{4} \ln t + 1$ , $x (1) = 0$
$t y^{'} = 2 y + t^{5} \sin (2 t) - t^{3} + 4 t^{4}$ , $y (π) = π^{4}$

Applications of First-Order Linear Equations

A brine solution flows at a constant rate of $4$ L/min into a large tank that initially held $100$ L of pure water. The solution inside the tank is kept well stirred and flows out of the tank at a rate of $3$ L/min. If the concentration of salt in the bring entering the tank is $0.2$ Kg/L, determine the mass of salt in the tank after $t$ min.

We must analyze two separate systems here. First, we have an input system where the rate of bring flowing into the tank is $4$ L/min, with a concentration of $0.2$ Kg/L. Let's calculate this input rate:

0.2 \frac{k g}{L} \times_{4} \frac{L}{m i n} = 0.8 \frac{k g}{m i n}

Now let's look at our output system. The amount of salt outputs at $3$ L/min, and the output system $A (t)$ is the amount of salt in the tank at any time $t$ , which will depend on the size of the tank.

\frac{A (t) k g}{100 L + t} \times 3 \frac{L}{m i n} = \frac{3 A (t) k g}{100 + t}

The DE is the difference between the input and the output systems we have defined above.

\frac{d A}{d t} = 0.8 - \frac{3 A (t)}{100 + t}

If we adjust this equation, we can turn this into a linear first order equation

\frac{d A}{d t} + \frac{3}{100 + t} A (t) = 0.8

where $P (t) = \frac{3}{100 + t}$ and $Q (t) = 0.8$

Now we can solve the DE using an integrating factor.

μ (t) = e^{\int \frac{3}{100 + t} d t} = e^{3 \ln | 100 + t |} = (100 + t)^{3}

A (t) = \frac{1}{μ (t)} [\int μ (t) Q (t) d t + C]

A (t) = \frac{1}{(100 + t)^{3}} [\int (100 + t)^{3} (0.8) d t + C]

A (t) = \frac{1}{(100 + t)^{3}} [\frac{0.8}{4} (100 + t)^{4} + C]

A (t) = \frac{1}{(100 + t)^{3}} [0.2 (100 + t)^{4} + C]

A (t) = 0.2 (100 + t) + \frac{C}{(100 + t)^{3}}

Thus the general solution is

A (t) = 0.2 (100 + t) + \frac{C}{(100 + t)^{3}}

We know that the tank starts with pure water, meaning the salt content at time $t = 0$ is $0$ , therefore $A (0) = 0$ .

A (0) = 0

0 = 0.2 (100 + 0) + \frac{C}{(100 + 0)^{3}}

0 = 0.2 (100) + \frac{C}{(100)^{3}}

C = - 20 \times (100)^{3}

C = - 2 \times 10^{7}

Therefore,

A (t) = 0.2 (100 + t) + \frac{- 2 \times 10^{7}}{(100 + t)^{3}}

Next Lesson: Lesson 7 - Exact Equations

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