Differential equations for NDA

Differential equations look advanced, but the NDA version is compact. The syllabus asks for the definition of order and degree, forming a differential equation from examples, general and particular solutions, first-order first-degree equations of various types, and applications to growth and decay. That list maps onto about five question patterns, and each has a routine you can learn in a week.

Order-and-degree questions are close to free marks if you know the one rule that trips people (clearing radicals first). Solving questions reward recognising the type in the first ten seconds.

Typical question types are:

• the order and degree of a given equation;
• the differential equation of a family of curves, or the number of arbitrary constants;
• solving a variables-separable equation;
• solving a homogeneous equation with y = vx;
• the integrating factor, or the full solution, of a linear equation;
• a particular solution that passes through a given point;
• growth and decay, such as a population doubling.

In this post, exp(u) means e raised to the power u. It keeps long exponents readable.

Order and degree

• Order is the order of the highest derivative in the equation.
• Degree is the power of that highest derivative, once the equation is a polynomial in all its derivatives.

The second condition matters. Before reading off the degree, clear any radicals or fractional powers on the derivatives. If a derivative sits inside sin, log or exp, the equation is not a polynomial in derivatives, and its degree is not defined.

Forming a differential equation

A family of curves with n arbitrary constants gives a differential equation of order n. Differentiate n times and eliminate the constants.

• y = Ae²ˣ (one constant): dy/dx = 2Ae²ˣ = 2y, so dy/dx = 2y.
• y = mx (lines through the origin): dy/dx = m = y/x, so x(dy/dx) = y.
• y = A cos x + B sin x (two constants): d²y/dx² = −y, so d²y/dx² + y = 0.
• y = A cos 2x + B sin 2x: d²y/dx² = −4y, so d²y/dx² + 4y = 0.
• x² + y² = r² (circles about the origin, one constant r): 2x + 2y(dy/dx) = 0, so x + y(dy/dx) = 0.

The reverse also holds: the general solution of an order-n equation has n arbitrary constants. A particular solution fixes them using given conditions, such as y = 3 when x = 0.

The three first-order types

1. Variables separable

If you can write the equation as g(y) dy = f(x) dx, integrate both sides.

dy/dx = xy gives dy/y = x dx, so ln|y| = x²/2 + C, which is y = C exp(x²/2).

2. Homogeneous

dy/dx = f(x, y) is homogeneous when f can be written as a function of y/x alone. Put y = vx, so dy/dx = v + x(dv/dx). The equation becomes separable in v and x.

3. Linear

The form is dy/dx + P(x)y = Q(x), where P and Q depend on x only.

• Integrating factor: IF = exp(∫P dx).
• Solution: y × IF = ∫(Q × IF) dx + C.

The logic is short. Multiplying by IF turns the left side into d/dx(y × IF), because d/dx of IF is P × IF. Then integrate both sides.

Useful IF values: P = 1 gives eˣ; P = 1/x gives x; P = 2/x gives x²; P = −1/x gives 1/x; P = tan x gives sec x.

Growth and decay

If a quantity changes at a rate proportional to itself, dN/dt = kN. Separating gives N = N₀ exp(kt), where N₀ is the value at t = 0. k > 0 means growth and k < 0 means decay.

• Doubling time = (ln 2)/k for growth.
• Half-life = (ln 2)/|k| for decay.
• The key fact: in equal time intervals, the quantity is multiplied by the same factor. If it doubles in 3 hours, it becomes 4 times in 6 hours and 8 times in 9 hours.

Worked NDA-style MCQs

Q1. The order and degree of [1 + (dy/dx)²] to the power 3/2 = d²y/dx² are:
(a) 2, 1 (b) 2, 2 (c) 1, 3 (d) 2, 3

Square both sides to clear the fractional power: (d²y/dx²)² appears, so the order is 2 and the degree is 2. Answer: (b).

Q2. The differential equation of y = A cos 2x + B sin 2x is:
(a) y″ − 4y = 0 (b) y″ + 2y = 0 (c) y″ + 4y = 0 (d) y′ + 4y = 0

y′ = −2A sin 2x + 2B cos 2x and y″ = −4A cos 2x − 4B sin 2x = −4y. Answer: (c).

Q3. The general solution of dy/dx = (1 + y²)/(1 + x²) is:
(a) tan⁻¹y − tan⁻¹x = C (b) tan⁻¹y + tan⁻¹x = C (c) y = x + C (d) ln(1 + y²) = ln(1 + x²) + C

Separate: dy/(1 + y²) = dx/(1 + x²). Integrating gives tan⁻¹y = tan⁻¹x + C. Answer: (a).

Q4. The solution of x(dy/dx) = x + y (for x > 0) is:
(a) y = x ln x + Cx (b) y = x + C (c) y = Cx² (d) y = ln x + C

This is homogeneous: dy/dx = 1 + y/x. With y = vx, v + x(dv/dx) = 1 + v, so dv = dx/x and v = ln x + C. Hence y = x ln x + Cx. Answer: (a).

Q5. The solution of dy/dx + y/x = x² is:
(a) y = x³/3 + C (b) xy = x⁴/4 + C (c) y = x⁴/4 + Cx (d) x²y = x⁵/5 + C

P = 1/x, so IF = exp(ln x) = x. Then xy = ∫x² × x dx = x⁴/4 + C. Answer: (b).

Q6. A bacterial culture grows at a rate proportional to its size and doubles in 3 hours. The time it takes to become 8 times its original size is:
(a) 6 hours (b) 8 hours (c) 9 hours (d) 24 hours

N = N₀ exp(kt) with exp(3k) = 2. We need exp(kt) = 8 = 2³ = exp(9k), so t = 9. Answer: (c).

Common mistakes

• Reading the degree before clearing radicals, or giving a degree when a derivative sits inside a sine or log.
• Confusing order with the number of constants. They are equal for the family's equation, but a question may ask either.
• Using the wrong IF because the equation was not first written as dy/dx + Py = Q. Divide by the coefficient of dy/dx before reading P.
• Dropping the constant in y × IF = ∫Q × IF dx + C. It changes the answer.
• Treating doubling as linear growth. Three doubling periods multiply the quantity by 2³ = 8, not by 1 + 3 = 4.

Practice set

1. The order and degree of y = x(dy/dx) + √(1 + (dy/dx)²) are: (a) 1, 1 (b) 1, 2 (c) 2, 1 (d) 2, 2
2. The general solution of a third-order differential equation contains how many arbitrary constants? (a) 1 (b) 2 (c) 3 (d) 0
3. The differential equation of all straight lines through the origin is: (a) dy/dx = y (b) x(dy/dx) = y (c) y(dy/dx) = x (d) dy/dx = x
4. The solution of dy/dx = eˣ⁺ʸ is: (a) eˣ + e⁻ʸ = C (b) eˣ − eʸ = C (c) eˣ + eʸ = C (d) e⁻ˣ + eʸ = C
5. The integrating factor of dy/dx + y tan x = sec x is: (a) cos x (b) sec x (c) tan x (d) exp(tan x)
6. The solution of x(dy/dx) = y is: (a) y = C (b) y = Cx (c) y = x + C (d) xy = C
7. If dy/dx = 2x and y = 3 when x = 0, then y at x = 2 is: (a) 4 (b) 5 (c) 7 (d) 11
8. The degree of sin(dy/dx) + y = 0 is: (a) 0 (b) 1 (c) 2 (d) not defined

Answers:

1. (b). Isolate the root and square: (y − x(dy/dx))² = 1 + (dy/dx)². The first derivative is the highest, and its highest power is 2.
2. (c). The order equals the number of constants.
3. (b). From y = mx, m = dy/dx = y/x.
4. (a). e⁻ʸ dy = eˣ dx gives −e⁻ʸ = eˣ + C, which rearranges to eˣ + e⁻ʸ = C.
5. (b). exp(∫tan x dx) = exp(ln sec x) = sec x.
6. (b). dy/y = dx/x, so ln y = ln x + C.
7. (c). y = x² + 3, so y(2) = 7.
8. (d). The derivative is inside a sine, so the equation is not a polynomial in dy/dx.

What to do next

• Make a one-line card for each type: how to spot it, the first move, one solved example.
• Solve ten order-and-degree questions from old papers and say the radical rule aloud each time.
• Revise indefinite integration if the integrals here slowed you down, and see definite integrals and area for the other half of integral calculus.