Differential Equations – AAI ATC Exam Prep

📘 About This Chapter

Differential Equations is a high-weightage topic in AAI ATC Mathematics. This chapter covers Order, Degree, General & Particular Solutions, methods like Variable Separable, Homogeneous, and Linear differential equations — all tested directly in AAI ATC written exams.

Order & Degree Variable Separable Homogeneous DE Linear DE Applications

Order and Degree of a Differential Equation

A differential equation is any equation that involves derivatives of a dependent variable with respect to an independent variable.

📌 Key Definitions

Order: The order of the highest-order derivative present in the equation.

Degree: The highest power of the highest-order derivative — only defined when the equation is a polynomial in its derivatives.

Example 1: Identify order and degree:

dy/dx = eˣ → Order = 1, Degree = 1

d²y/dx² + y = 0 → Order = 2, Degree = 1

dy/dx + sin(dy/dx) = 0 → Order = 1, Degree = NOT DEFINED

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⚠️ Important Note

Order and degree (if defined) of a differential equation are always positive integers. Degree is undefined when a transcendental function (sin, cos, log, e) of the derivative is present.

Practice MCQs – Order & Degree

Q1. The order and degree of the differential equation (d²y/dx²)³ + (dy/dx)⁴ + y⁵ = 0 are respectively:

✅ Answer: B — Highest derivative is d²y/dx² (order = 2). Its highest power is 3 (degree = 3).

Q2. The degree of the differential equation d⁴y/dx⁴ + sin(y''') = 0 is:

✅ Answer: D — sin(y''') makes this a non-polynomial equation in derivatives. So degree is NOT DEFINED.

Q3. The order of the equation 2x² (d²y/dx²) – 3(dy/dx) + y = 0 is:

✅ Answer: A — The highest order derivative is d²y/dx², so order = 2.

General and Particular Solutions

Unlike algebraic equations (whose solutions are numbers), the solution of a differential equation is a function.

📌 Key Definitions

General Solution: Contains arbitrary constants equal to the order of the DE. Example: y = a sin(x + b)

Particular Solution: Obtained by giving specific values to the arbitrary constants using initial/boundary conditions.

Example: Verify y = e⁻³ˣ is a solution of y'' + dy/dx − 6y = 0

y = e⁻³ˣ → dy/dx = −3e⁻³ˣ → d²y/dx² = 9e⁻³ˣ

LHS = 9e⁻³ˣ + (−3e⁻³ˣ) − 6e⁻³ˣ = 9e⁻³ˣ − 9e⁻³ˣ = 0 = RHS ✓

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Number of arbitrary constants in general solution = order of the differential equation.

Particular solution has zero arbitrary constants.

Practice MCQs – General & Particular Solutions

Q1. The number of arbitrary constants in the general solution of a 4th-order differential equation is:

✅ Answer: D — The general solution of an nth-order DE contains exactly n arbitrary constants.

Q2. Verify: y = A·eˣ is a solution of dy/dx − y = 0. The value of A when y = 3 at x = 0 is:

✅ Answer: B — y(0) = A·e⁰ = A = 3, so A = 3. This gives the particular solution y = 3eˣ.

Q3. The number of arbitrary constants in the particular solution of a 3rd-order DE is:

✅ Answer: D — A particular solution is free from arbitrary constants. It's obtained by substituting specific values.

Variable Separable Method

Used when the DE can be written such that all y-terms are on one side with dy and all x-terms on the other side with dx.

📌 Method

If dy/dx = h(y)·g(x), then separate and integrate:

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

∫ [1/h(y)] dy = ∫ g(x) dx + C

Example: Find general solution of dy/dx = (x+1)/(2−y)

Step 1: Separate → (2−y)dy = (x+1)dx

Step 2: Integrate → 2y − y²/2 = x²/2 + x + C₁

Step 3: Rearrange → x² + y² + 2x − 4y + C = 0

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Application (AAI ATC Favourite): In a bank, principal P increases at 5% per year. In how many years does ₹1000 double?

dP/dt = P/20 → Integrate → P = C·e^(t/20)

At t=0: C=1000 → 2000 = 1000·e^(t/20) → t = 20·ln(2)

Practice MCQs – Variable Separable

Q1. The general solution of dy/dx = (1+y²)/(1+x²) is:

✅ Answer: B — Separate: dy/(1+y²) = dx/(1+x²). Integrate both: tan⁻¹y = tan⁻¹x + C.

Q2. The general solution of dy/dx = e^(x+y) is:

✅ Answer: A — e^(x+y) = eˣ·eʸ. So e⁻ʸ dy = eˣ dx → −e⁻ʸ = eˣ + C₁ → eˣ + e⁻ʸ = C.

Q3. A principal ₹100 doubles in 10 years at continuous compounding. The rate r% per year satisfies (log₂ = 0.6931):

✅ Answer: C — dP/dt = rP/100. Solution: P = P₀·e^(rt/100). 2P₀ = P₀·e^(r·10/100) → r = 10·ln2 ≈ 6.931%.

Homogeneous Differential Equations

A function F(x, y) is homogeneous of degree n if F(λx, λy) = λⁿ·F(x, y) for any nonzero constant λ.

A DE dy/dx = F(x, y) is homogeneous if F(x, y) is a homogeneous function of degree zero.

📌 Method of Solution

Substitute y = vx (and dy/dx = v + x·dv/dx) to convert to variable separable form.

dy/dx = g(y/x) → Put y = vx → x·dv/dx = g(v) − v

Separate → dv / [g(v)−v] = dx/x → Integrate both sides

After integration, replace v by y/x

Example: Show (x−y)dy/dx = x+2y is homogeneous and solve.

dy/dx = (x+2y)/(x−y) → F(λx,λy) = λ(x+2y)/λ(x−y) = λ⁰·F → Degree 0 ✓

Put y = vx: v + x·dv/dx = (1+2v)/(1−v)

General solution: log|x²+xy+y²| = 2√3·tan⁻¹[(x+2y)/(√3·x)] + C

🖼️ Image Space
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⚠️ Special Case

If the homogeneous DE is of the form dx/dy = h(x/y), substitute x = vy instead.

Practice MCQs – Homogeneous Differential Equations

Q1. Which of the following is a homogeneous function of degree 2?

✅ Answer: B — F(λx, λy) = λ²y² + 2λ²xy = λ²(y²+2xy) = λ²·F(x,y). So degree = 2.

Q2. For homogeneous DE dy/dx = (x²+y²)/(2xy), the substitution to use is:

✅ Answer: B — Standard substitution for homogeneous DE of the form dy/dx = f(y/x) is y = vx.

Q3. The general solution of (x²−y²)dx + 2xy·dy = 0 is:

✅ Answer: C — After substitution y = vx, the general solution is x² + y² = Cy (constant C = ±C₁).

Linear Differential Equations (First Order)

A first order linear DE has the standard form:

dy/dx + P(x)·y = Q(x)

Where P and Q are functions of x only (or constants).

📌 Solution Steps

Step 1: Identify P(x) and Q(x)

Step 2: Find I.F. = e^∫P dx

Step 3: Write solution as:

y × (I.F.) = ∫ Q × (I.F.) dx + C

Example: Solve dy/dx − y = cos x

P = −1, Q = cos x → I.F. = e^∫(−1)dx = e⁻ˣ

y·e⁻ˣ = ∫ e⁻ˣ·cos x dx + C

Solving integral: I = e⁻ˣ(sin x − cos x)/2

∴ y = (sin x − cos x)/2 + C·eˣ

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Add a clear flowchart: "How to identify and solve a linear DE" — with boxes for Standard form → Find P, Q → Compute I.F. → Write solution

⚠️ Alternative Form (in y)

dx/dy + P₁(y)·x = Q₁(y)

Then I.F. = e^∫P₁ dy, and solution: x·(I.F.) = ∫ Q₁·(I.F.) dy + C

Method	Condition	Substitution / Key Step
Variable Separable	dy/dx = g(x)·h(y)	Separate x and y, then integrate
Homogeneous	F(x,y) of degree 0	y = vx or x = vy
Linear DE	dy/dx + Py = Q	I.F. = e^∫P dx

Practice MCQs – Linear Differential Equations

Q1. The integrating factor of dy/dx + (2/x)·y = x² is:

✅ Answer: B — P = 2/x. I.F. = e^∫(2/x)dx = e^(2ln|x|) = e^(ln x²) = x².

Q2. The integrating factor of the linear DE (1−y²)dx/dy + yx = ay (−1 < y < 1) is:

✅ Answer: D — Standard form: dx/dy + y/(1−y²)·x = ay/(1−y²). P₁ = y/(1−y²). I.F. = e^∫y/(1−y²)dy = 1/√(1−y²).

Q3. The general solution of eˣ dy + (yeˣ + 2x) dx = 0 is:

✅ Answer: C — Rewrite: dy/dx + y = −2xe⁻ˣ. I.F. = eˣ. Solution: d(yeˣ)/dx = −2x → yeˣ = −x² + C → yeˣ + x² = C.

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