Differential Equations

A differential equation is an equation that relates a function to its derivatives. This chapter is about the methods used to solve these equations, which means finding the original function that satisfies the relationship. You will begin by learning to classify equations by their order and degree, and also how to form a differential equation from a given family of curves by eliminating arbitrary constants.

The main part of the chapter covers techniques for solving first-order, first-degree differential equations. The three principal methods are the variables separable method, where terms are grouped by variable before integration; the method for homogeneous equations, which uses the substitution y = vx; and the procedure for linear differential equations, which requires finding an integrating factor. You will also learn to recognize and solve equations that can be reduced to one of these standard forms through substitution.

The most common difficulty is identifying the correct solution method for a given equation, as it may require algebraic manipulation to fit a standard form. Application problems, which ask you to model a situation with a differential equation, also require careful setup. This chapter is entirely dependent on your knowledge of integration. A thorough understanding of all integration techniques is essential to solve the problems you will encounter.

Formulas for Differential Equations

Variables Separable Form: $f(y) \, dy = g(x) \, dx$ The solution is found by integrating both sides: ∫f(y) dy = ∫g(x) dx + C.
Homogeneous Equation Form: $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$ Solved by substituting y = vx, which leads to a variables separable equation.
Linear Differential Equation (First Order): $\frac{dy}{dx} + P(x)y = Q(x)$ This is the standard form for a linear equation in y.
Integrating Factor (I.F.): $\text{I.F.} = e^{\int P(x) \, dx}$ Calculated for a linear differential equation in the standard form.
Solution of Linear DE: $y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + C$ This gives the general solution for the linear differential equation.
Bernoulli's Equation: $\frac{dy}{dx} + P(x)y = Q(x)y^n$ This is reducible to a linear form for n ≠ 0 and n ≠ 1.
Substitution for Bernoulli's Equation: $z = y^{1-n}$ This substitution converts Bernoulli's equation into a linear DE in z.
Linear Differential Equation in x: $\frac{dx}{dy} + P(y)x = Q(y)$ An alternative form where roles of x and y are interchanged.
Integrating Factor for Linear DE in x: $\text{I.F.} = e^{\int P(y) \, dy}$ Calculated for a linear differential equation in x.

One of these questions, worked through

JEE Advanced 2018

Let $f:[0,\infty) \to \mathbb{R}$ be a continuous function such that $f(x) = 1 - 2x + \int_0^x e^{x-t} f(t)\,dt$ for all $x \in [0,\infty)$ . Then, which of the following statement(s) is(are) TRUE?

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Let f:[0,∞) → R be a continuous function such that f(x) = 1 - 2x + ∫₀ˣ eˣ⁻ᵗ f(t) dt for all x ∈ [0,∞). Then, which of the following statement(s) is(are) TRUE? $\text{Given: } f(x) = 1 - 2x + \displaystyle\int_0^x e^{x-t} f(t)\,dt,\quad f \text{ continuous on } [0,\infty)$ $\text{To find: which of (1)--(4) are TRUE}$ $1.\ \text{differentiate to get an ODE}$ $2.\ \text{solve for } f(x)$ $3.\ \text{check points, then the area}$ $f(x) = 1 - 2x + e^x \displaystyle\int_0^x e^{-t} f(t)\,dt$ $f'(x) = -2 + e^x\displaystyle\int_0^x e^{-t}f(t)\,dt + e^x\cdot e^{-x}f(x)$ $f'(x) = -2 + (f(x)-1+2x) + f(x) = 2f(x) + 2x - 3$ $f(0) = 1$ $f' - 2f = 2x - 3,\qquad \text{IF} = e^{-2x}$ $\dfrac{d}{dx}\left(f e^{-2x}\right) = (2x-3)e^{-2x}$ $f e^{-2x} = e^{-2x}(1-x) + C$ $x=0:\ 1 = 1 + C \Rightarrow C = 0$ $f(x) = 1 - x$ $f(1) = 0 \neq 2 \Rightarrow \text{(1) FALSE}$ $f(2) = -1 \Rightarrow \text{(2) TRUE}$ $A = \displaystyle\int_0^1 \left(\sqrt{1-x^2} - (1-x)\right)dx$ $A = \dfrac{\pi}{4} - \dfrac12 = \dfrac{\pi-2}{4} \Rightarrow \text{(3) TRUE},\ \text{(4) FALSE}$ $\boxed{\text{Answer: (2) and (3)}}$

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Past-paper questions on Differential Equations

JEE Advanced 21 questions