Differential Equations

An equation that involves an independent variable, a dependent variable, and one or more derivatives of the dependent variable with respect to the independent variable is called a differential equation (DE). When there is only one independent variable, it is called an ordinary differential equation (ODE) — this is the only type studied in this chapter.

Examples of differential equations: dy/dx = cos x, d²y/dx² + y = 0, and (dy/dx)³ + 2y = x. In each case, the equation involves derivatives such as dy/dx, d²y/dx², etc., alongside x and/or y.

The order of a differential equation is the order of the highest derivative appearing in it. The degree is the power (positive integer exponent) of the highest order derivative, provided the differential equation is expressible as a polynomial in all the derivatives that occur in it — that is, it must be free of radicals and fractional powers of derivatives, and free of derivatives inside transcendental functions such as sin, cos, log, or e. If the equation cannot be written in this polynomial form, the degree is not defined.

Worked Example 1: Find the order and degree of d²y/dx² + (dy/dx)³ + y = 0.
The highest order derivative is d²y/dx², so the order is 2. This derivative appears to the power 1 (only (dy/dx) is cubed, not the second derivative), so the degree is 1.

Worked Example 2: Find the order and degree of √(1 + (dy/dx)²) = d²y/dx².
This is not yet a polynomial in the derivatives because of the square root. Squaring both sides: 1 + (dy/dx)² = (d²y/dx²)². Now it is a polynomial in the derivatives. The highest order derivative is d²y/dx² (order 2), raised to the power 2, so the degree is 2.

Worked Example 3: Find the order and degree of (d³y/dx³)² + cos(dy/dx) = 0.
The highest derivative is d³y/dx³, so the order is 3. However, the term cos(dy/dx) means the equation is not a polynomial in dy/dx (no amount of algebra removes the cosine), so the degree is not defined.

A function y = f(x) that satisfies a differential equation (i.e., substituting it and its derivatives makes the equation an identity) is called a solution of that differential equation. The general solution is a solution that contains as many independent arbitrary constants as the order of the differential equation. When specific numerical values are assigned to these constants — usually determined from given initial or boundary conditions — the result is called a particular solution.

To verify that a given function is a solution, differentiate it the required number of times and substitute y, y′, y″, etc. into the differential equation to check the equation is satisfied identically.

Worked Example: (a) Verify that y = a cos x + b sin x (a, b arbitrary constants) is a solution of y″ + y = 0. (b) The related equation y″ − 4y = 0 has general solution y = A e^2x + B e^−2x; find the particular solution satisfying y(0) = 0 and y′(0) = 4.

First, for y = a cos x + b sin x: y′ = −a sin x + b cos x, and y″ = −a cos x − b sin x = −y. So y″ + y = −y + y = 0, which verifies the solution for every choice of a and b — this is the general solution since it has 2 arbitrary constants matching the order 2 of the equation.

Now consider the second-order DE y″ − 4y = 0, whose general solution is y = A e^2x + B e^−2x. To find the particular solution satisfying y(0) = 0 and y′(0) = 4: differentiate to get y′ = 2A e^2x − 2B e^−2x. Substituting x = 0: A + B = 0, and 2A − 2B = 4, i.e., A − B = 2. Solving these two equations simultaneously (using B = −A): A − (−A) = 2 ⇒ 2A = 2 ⇒ A = 1, B = −1. So the particular solution is y = e^2x − e^−2x.

Every family of curves depending on n independent arbitrary constants satisfies a differential equation of order n, obtained by eliminating the constants. The procedure is: (i) differentiate the given relation n times with respect to x, obtaining n additional equations; (ii) use the original relation together with these n derivative-equations to eliminate all n constants, leaving one equation connecting x, y, and derivatives of y up to order n.

Worked Example 1 (one constant): Form the differential equation representing the family of straight lines through the origin, y = mx.
Differentiating once: dy/dx = m. Since the family has only one arbitrary constant (m), one differentiation suffices. Substitute m = dy/dx back into y = mx: y = x(dy/dx), i.e., x(dy/dx) − y = 0, a first-order DE.

Worked Example 2 (one constant): Form the differential equation of the family of circles centred at the origin, x² + y² = a².
Differentiating with respect to x: 2x + 2y(dy/dx) = 0, which simplifies directly to x + y(dy/dx) = 0 — the constant a has already disappeared, so no further substitution is needed.

Worked Example 3 (two constants): Form the differential equation representing y = A e^3x + B e^−3x.
Since there are two arbitrary constants, differentiate twice. y′ = 3A e^3x − 3B e^−3x. y″ = 9A e^3x + 9B e^−3x = 9(A e^3x + B e^−3x) = 9y. So the required differential equation is d²y/dx² − 9y = 0, of order 2 (matching the two constants eliminated).

A first-order, first-degree differential equation of the form dy/dx = f(x)·g(y) is said to have variables separable, because it can be rearranged so that all terms in y (with dy) appear on one side and all terms in x (with dx) appear on the other:

dy/g(y) = f(x) dx

Integrating both sides gives the general solution: ∫ dy/g(y) = ∫ f(x) dx + C, where C is the arbitrary constant of integration.

Worked Example 1 (general solution): Solve dy/dx = (1 + y²)/(1 + x²).
Separating variables: dy/(1 + y²) = dx/(1 + x²). Integrating both sides: tan⁻¹y = tan⁻¹x + C. This is the required general solution.

Worked Example 2 (particular solution): Solve dy/dx = 2xy given that y = 1 when x = 0.
Separating variables: dy/y = 2x dx (assuming y ≠ 0). Integrating: ln|y| = x² + C₁, so y = e^x²+C₁ = k e^x², where k = e^C₁ is a new arbitrary constant. Applying the condition y(0) = 1: 1 = k·e⁰ = k, so k = 1. Hence the particular solution is y = e^x². (Check: dy/dx = 2x e^x² = 2xy ✓, and y(0) = e⁰ = 1 ✓.)

A function F(x, y) is called a homogeneous function of degree n if F(λx, λy) = λⁿF(x, y) for a nonzero constant λ. A differential equation dy/dx = F(x, y) is called homogeneous if F(x, y) is a homogeneous function of degree zero — equivalently, it can always be written in the form dy/dx = g(y/x), depending only on the ratio y/x (or x/y).

To solve a homogeneous DE, substitute y = vx, treating v as a new function of x. Then dy/dx = v + x(dv/dx). Substituting this into the equation converts it into a variable-separable equation in v and x, which can then be solved by the method of the previous section, and finally v is replaced back by y/x.

Worked Example: Solve (x² + y²) dx − 2xy dy = 0.
Rewrite as dy/dx = (x² + y²)/(2xy). The right side is homogeneous of degree zero (numerator and denominator are both degree 2), so substitute y = vx, dy/dx = v + x(dv/dx):

v + x(dv/dx) = (x² + v²x²)/(2x·vx) = (1 + v²)/(2v)

So x(dv/dx) = (1 + v²)/(2v) − v = (1 + v² − 2v²)/(2v) = (1 − v²)/(2v). Separating variables: 2v/(1 − v²) dv = dx/x. Integrating the left side (let u = 1 − v², du = −2v dv, so the integral becomes −∫du/u = −ln|u|): −ln|1 − v²| = ln|x| + C₁, which rearranges to ln|x(1 − v²)| = −C₁, i.e., x(1 − v²) = C (absorbing the sign and exponentiating into a single constant C). Substituting back v = y/x: x(1 − y²/x²) = C ⇒ x − y²/x = C. Multiplying through by x gives the general solution x² − y² = Cx.

A differential equation of the form dy/dx + Py = Q, where P and Q are functions of x alone (or constants), is called a linear differential equation in y. It is solved using the integrating factor (I.F.), defined as I.F. = e^{∫P dx}. Multiplying both sides of the equation by the I.F. makes the left-hand side an exact derivative of the product y·(I.F.), so that the general solution is:

y · (I.F.) = ∫ Q · (I.F.) dx + C

Symmetrically, an equation of the form dx/dy + Px = Q, with P, Q functions of y alone, is linear in x, with I.F. = e^{∫P dy} and solution x·(I.F.) = ∫Q·(I.F.) dy + C. This form is useful whenever it is easier to treat x as the dependent variable (for example, when Q would otherwise be difficult to express as a function of x).

Worked Example: Solve dy/dx + y/x = x², for x > 0.
Here P = 1/x and Q = x². The integrating factor is I.F. = e^∫(1/x)dx = e^{ln x} = x (since x > 0). Multiplying the equation by x: x(dy/dx) + y = x³, and the left side is exactly d/dx(xy). So d/dx(xy) = x³. Integrating both sides with respect to x: xy = x⁴/4 + C. Hence the general solution is xy = x⁴/4 + C, or equivalently y = x³/4 + C/x.