# Linear Differential Equations

fact

A linear first-order differential equation is a differential equation that can be placed in the form:
$$\frac{dy}{dt} + p(t)y = g(t)$$
Where both \(p(t)\) and \(g(t)\) are continuous functions.

fact

The solution to a first-order differential equation is given by:
$$y = \frac{\int \mu (t)g(t)dt + c}{\mu (t)}$$
Where $$\mu (t) = e^{\int p(t)dt}$$
And \(c\) is the integration constant from the integration in the numerator. We write it explicitly because it's very important to include and is incredibly common to forget about.

example

Solve \(y' = 9.8 - 0.196t\)

Rearrange to get it into canonical form: \(y' + 0.196t = 9.8\) Here \(p(t) = 0.196\) and \(g(t) = 9.8\) So \(\mu = e^{\int 0.196dt} = e^{0.196t}\) Now \(y = e^{-0.196t}(\int 9.8e^{0.196t}dt + c) = 50 + ce^{-0.196t}\)example

Solve \(xy' + 2y = x^2 - x + 1\)

Rearranging into linear canonical form by dividing everything by \(x\) we get: \(y' + \frac{2}{x}y = x - 1 + \frac{1}{x}\) So \(p(x) = \frac{2}{x}\) and \(g(x) = x - 1 + \frac{1}{x}\) Then \(\mu = e^{\int \frac{2}{x}dx} = e^{2\ln(x)} = e^{\ln(x^2)} = x^2\) And \(y = x^{-2}(\int x^2\times (x - 1 + \frac{1}{x})dx + c)\) \( y = x^{-2}(\int (x^3 - x^2 + x)dx + c) \) \( y = x^{-2}(\frac{x^4}{4} - \frac{x^3}{3} + \frac{x^2}{2} + c) \) \( y = \frac{x^2}{4} - \frac{x}{3} + \frac{1}{2} + \frac{c}{x^2} \)example

Solve \(y' - y - xe^x = 0\)

Rewrite into standard form: \(y' - y = xe^x\) Now \(p(x) = -1\) and \(g(x) = xe^x\) So \(\mu = e^{\int -1dx} = e^{-x}\) And \(y = e^x(\int(e^{-x}xe^xdx) + c)\) \( y = e^x(\frac{1}{2}x^2 + c) \)
practice problems