Differential Equations

🎯 Key Points

• Order = highest derivative present; Degree = power of that highest derivative (only defined when the equation is polynomial in its derivatives)
• Linear first-order dy/dx+P(x)y=Q(x): integrating factor IF=e^∫P dx, solution y·IF=∫Q·IF dx — memorize this exact template, it solves the entire category
• Number of arbitrary constants in the general solution = order of the DE; a particular solution fixes those constants via given initial conditions
• To FORM a DE from a family of curves with n constants: differentiate n times, then eliminate the constants between the original relation and its derivatives

Differential Equations

A differential equation (DE) contains derivatives. Solutions describe how quantities change over time or position.

Order and Degree

• Order = highest derivative in the equation
• Degree = power of the highest derivative (when polynomial)
• Example: d²y/dx² + 3(dy/dx)² + y = 0 has order 2, degree 1

First Order DEs

• Separable: dy/dx = f(x)g(y) → separate variables: dy/g(y) = f(x)dx, then integrate both sides
• Linear: dy/dx + P(x)y = Q(x). Integrating factor = e^∫P dx. Solution: y × IF = ∫Q × IF dx
• Homogeneous: Put y = vx, then dy/dx = v + x(dv/dx)

Second Order Linear DEs

For ay'' + by' + cy = 0:

• Characteristic equation: am² + bm + c = 0
• Two distinct real roots (m₁, m₂): y = Ae^(m₁x) + Be^(m₂x)
• Equal roots (m₁ = m₂ = m): y = (A + Bx)e^(mx)
• Complex roots (m = α ± iβ): y = e^(αx)(A cosβx + B sinβx)

Applications

• Exponential growth/decay: dy/dt = ky → y = y₀e^(kt)
• Newton's law of cooling: dT/dt = -k(T - T₀)
• Simple harmonic motion: d²x/dt² = -ω²x
• Population models, radioactive decay, charging capacitors

Formation of a Differential Equation

A differential equation can be formed from a family of curves by eliminating the arbitrary constants. If a relation has n arbitrary constants, differentiate it n times and eliminate the constants between the original relation and its derivatives to get an equation of order n.

Worked example: Form the DE for the family y = Ax^2 (A is an arbitrary constant).

Differentiating: dy/dx = 2Ax. From the original equation, A = y/x^2, so dy/dx = 2x(y/x^2) = 2y/x, which gives the DE x(dy/dx) = 2y, a first-order equation with the constant eliminated.

Worked Example: Variable Separable

Solve dy/dx = 2xy with the initial condition y(0) = 1.

Separate variables: dy/y = 2x dx. Integrating both sides: ln|y| = x^2 + C, so y = A e^(x^2) where A = e^C.

Applying y(0) = 1: 1 = A e^0 = A, so A = 1. The particular solution is y = e^(x^2).

Worked Example: Linear Differential Equation

Solve dy/dx + y/x = x (for x > 0).

This is in the standard linear form dy/dx + P(x)y = Q(x) with P(x) = 1/x and Q(x) = x.

Integrating factor: IF = e^∫(1/x)dx = e^(ln x) = x.

Solution: y x IF = ∫Q x IF dx, so y x x = ∫x x x dx = ∫x^2 dx = x^3/3 + C.

Therefore y = x^2/3 + C/x. This general solution contains one arbitrary constant, consistent with a first-order equation.

General Solution vs Particular Solution

• A general solution contains as many independent arbitrary constants as the order of the differential equation.
• A particular solution is obtained by assigning specific values to the constants, usually using given initial conditions (initial value problem) such as y(x0) = y0.
• Geometrically, the general solution represents a family of curves, while a particular solution picks out exactly one curve from that family.

Verifying a Solution of a Differential Equation

A function y = φ(x) is a solution of a differential equation if it, together with its derivatives, satisfies the equation identically for all x in the domain.

• Method: compute the required derivatives of the given function, substitute them into the differential equation, and check that the two sides become equal.
• Example: verify that y = e^(-x) + 1 is a solution of y'' + y' = 0. Here y' = -e^(-x) and y'' = e^(-x), so y'' + y' = e^(-x) - e^(-x) = 0, which satisfies the equation.
• A relation defining y implicitly can also be a solution; differentiate implicitly and substitute in the same way.

Homogeneous Differential Equations

A first-order equation dy/dx = F(x, y) is homogeneous if F can be written purely as a function of the ratio y/x, i.e. dy/dx = g(y/x). Equivalently it has the form dy/dx = f(x, y)/h(x, y) where f and h are homogeneous functions of the same degree.

• Substitution: put y = vx, so dy/dx = v + x(dv/dx). This reduces the equation to a variables-separable form in v and x.
• After separating and integrating, replace v by y/x to return to the original variables.
• If the equation is more naturally a function of x/y, use the symmetric substitution x = vy with dx/dy = v + y(dv/dy) instead.
• Example: for dy/dx = (x + y)/x = 1 + y/x, put y = vx to get v + x(dv/dx) = 1 + v, so x(dv/dx) = 1, giving dv = dx/x, hence v = ln|x| + C and y = x(ln|x| + C).

🚀 JEE Advanced Edge

Recognizing a "disguised" homogeneous or linear equation: An equation like dy/dx = (x+y)/(x-y) doesn't look separable, but dividing numerator and denominator by x reveals it depends only on the ratio y/x — the signature of a homogeneous equation, solved via y=vx. Many JEE differential equations require this kind of algebraic rearrangement BEFORE the standard method (separable, linear, or homogeneous) becomes visible.

Orthogonal trajectories: Two families of curves are orthogonal trajectories of each other if every curve in one family intersects every curve in the other family at right angles — found by taking the differential equation of one family, replacing dy/dx with -dx/dy (the perpendicular-slope condition), and solving the resulting DE to get the second family. This technique connects differential equations directly back to the perpendicular-slope idea from coordinate geometry.

Worked problem: Solve the differential equation dy/dx = y/x + tan(y/x) (a homogeneous equation in disguise). Approach: Substitute y=vx, so dy/dx=v+x(dv/dx). The equation becomes v+x(dv/dx) = v+tan(v), i.e. x(dv/dx)=tan(v). Separating: cot(v)dv = dx/x. Integrating: ln|sin v| = ln|x|+C, so sin(v) = Ax. Substituting back v=y/x: sin(y/x) = Ax is the general solution.