2.6: First Order Linear Differential Equations
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- 381
In this section we will concentrate on first order linear differential equations. Recall that this means that only a first derivative appears in the differential equation and that the equation is linear. The general first order linear differential equation has the form
\[ y' + p(x)y = g(x) \]
Before we come up with the general solution we will work out the specific example
\[ y' + \frac{2}{x y} = \ln \, x. \]
The strategy for solving this is to realize that the left hand side looks a little like the product rule for differentiation. The product rule is
\[ (my)' = my' + m'y. \]
This leads us to multiplying both sides of the equation by m which is called an integrating factor.
\[ m y' + m \frac{2}{x y} = m \ln\, x. \]
We now search for a \(m\) with
\[ m' = m \frac{2}{x} \]
or
\[ \frac{dm}{m} = \frac{2}{x} \; dx. \]
Integrating both sides, produces
\[ \ln\, m = 2\ln\, x = \ln(x^2) \]
or \( m = x^2 \) by exponentiating both sides.
Going back to the original differential equation and multiplying both sides by \( x^2 \), we get
\[ x^2y' + 2xy = x^2 \ln\, x . \]
Using the product rule in reverse gives
\[ (x^2y)' = x^2 \ln \, x . \]
Now integrate both sides. Note that the integral of the derivative is the original. Integrate by parts to get
\[ \int x^2 \, \ln \, x \, dx. \]
- \(u = \ln x\) and \(dv = x^2 \,dx\)
- \(du = \dfrac{1}{x} dx \) and \( v = \dfrac{1}{3} x^3\)
\[ = \dfrac{x^3 \ln\, x}{3} - \int \dfrac{x^2}{3} \, dx = \dfrac{x^3 \ln \, x}{3} - \dfrac{x^3}{9} +C \]
Hence
\[ x^2y = \frac{1}{3} x^3 \ln x - \frac{1}{9} x^3 + C. \]
Divide by \(x^2\)
\[ y = \dfrac{1}{3} x \ln \, x - \dfrac{1}{9} x + \dfrac{C}{x^2} . \]
Notice that when \(C\) is nonzero, the solutions are undefined at \(x = 0\). Also given an initial value with \(x\) positive, there will be no solution for negative \(x\). Now we will derive the general solution to first order linear differential equations.
Consider
\[ y' + p(t)y = g(t). \]
We multiply both sides by \(m\) to get
\[ my' + mp(x)y = mg(x). \]
We now search for an \(m\) with
\[ m' = mp(x) \]
or
\[ \dfrac{dm}{m} = p(x) \, dx . \]
Integrating both sides, produces
\[ \text{ln} \, m = \int p(x)\, dx \]
exponentiating both sides
\[ m = e^{\int p(x)\, dx}. \]
Going back to the original differential equation and multiplying both sides by \(m\), we get
\[ my' + mp(x)y = mg(x) \]
\[ (my)' = mg(x) \]
\[ my = \int \mu \, g(x) \, dx. \]
Solving for \(y\) gives
\[ y = e^{-\int p(x)\, dx}\int g(x) e^{\int p(x) \, dx } \, dx .\]