Linear Differential Equation

A differential equation of the type dy/dx + Py = Q, where P and Q are functions of x (or constants), is called a LINEAR DIFFERENTIAL EQUATION. Its distinguishing feature is that the dependent variable y — and its derivative — appear only to the first power and are never multiplied together or by another function of y.

To solve it, both sides are multiplied by e^(∫P dx), called the INTEGRATING FACTOR (I.F.), giving e^(∫Pdx)(dy/dx) + Py·e^(∫Pdx) = Q·e^(∫Pdx). The key observation is that the left-hand side is exactly the derivative, with respect to x, of the product y·e^(∫Pdx) — because d/dx[e^(∫Pdx)] = P·e^(∫Pdx) by the fundamental theorem of calculus, the product rule applied to y·e^(∫Pdx) reproduces precisely the left side above. So the equation collapses to d/dx[y·e^(∫Pdx)] = Q·e^(∫Pdx), and integrating both sides with respect to x gives the general solution formula: y·e^(∫Pdx) = ∫Q·(e^(∫Pdx))dx + c, i.e. y·(I.F.) = ∫Q·(I.F.)dx + c, where I.F. = e^(∫Pdx) is the integrating factor.

A parallel note records the mirror-image case: for the linear differential equation dx/dy + Px = Q (here P, Q are functions of y, or constants — the equation is linear in x as a function of y instead), the general solution is x·(I.F.) = ∫Q·(I.F.)dy + c, where I.F. = e^(∫Pdy).

Three solved examples apply the dy/dx-form recipe, and a fourth is a genuine word problem:

Ex.1(i) dy/dx + y = e^(−x): here P=1, Q=e^(−x), so I.F. = e^(∫dx) = e^x. Then y·e^x = ∫e^(−x)·e^x dx + c = ∫e⁰dx + c = ∫dx + c = x + c, i.e. y·e^x = x + c, the general solution.

Ex.1(ii) x sin x(dy/dx) + (x cos x + sin x)y = sin x: dividing by x sin x gives dy/dx + [cot x + 1/x]y = 1/x, so P = cot x + 1/x, Q = 1/x. I.F. = e^(∫(cot x + 1/x)dx) = e^(log|sin x| + log x) = x sin x. Then x sin x·y = ∫(1/x)(x sin x)dx + c = ∫sin x dx + c = −cos x + c, i.e. xy·sin x + cos x = c.

Ex.1(iii) (1+y²)dx = (tan⁻¹y − x)dy: rewriting in dx/dy form, dx/dy = (tan⁻¹y − x)/(1+y²), i.e. dx/dy + x/(1+y²) = tan⁻¹y/(1+y²) — linear in x, with P = 1/(1+y²), Q = tan⁻¹y/(1+y²). I.F. = e^(∫dy/(1+y²)) = e^(tan⁻¹y). Then x·e^(tan⁻¹y) = ∫[tan⁻¹y/(1+y²)]·e^(tan⁻¹y)dy + c; substituting t = tan⁻¹y (so dy/(1+y²) = dt) turns the integral into ∫t·eᵗdt = t·eᵗ − eᵗ + c (by parts), so x·e^(tan⁻¹y) = tan⁻¹y·e^(tan⁻¹y) − e^(tan⁻¹y) + c, i.e. x + 1 − tan⁻¹y = c·e^(−tan⁻¹y), the solution. …