Linear Differential Equations

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.

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.

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 \(y = e^{-0.196t}(\int 9.8e^{0.196t}dt + c) = 50 + ce^{-0.196t}\)

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 \(y = x^{-2}(\int x^2\times (x - 1 + \frac{1}{x})dx + c)\) \( y = x^{-2}(\int (x^3 - x^2 + x)dx + c) \) \( y = x^{-2}(\frac{x^4}{4} - \frac{x^3}{3} + \frac{x^2}{2} + c) \) \( y = \frac{x^2}{4} - \frac{x}{3} + \frac{1}{2} + \frac{c}{x^2} \)

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 \(y = e^x(\int(e^{-x}xe^xdx) + c)\) \( y = e^x(\frac{1}{2}x^2 + c) \)