Differential Equations — Class 12 (Advanced)

Introduction — At a glance

These notes cover the full Class 12 chapter on Differential Equations. You will find concise theorems, precise formula tables, step-by-step proofs and 8 solved examples that transfer the chapter to memory — the drills, diagrams (SVG) and mnemonics help long-term recall.

Why this page?

• Exam-focused: board & competitive exam relevance (SSC, Railway, Police Bharti, NEET quick-revision).
• Memory-first layout: theorem → proof → example → mnemonic.
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Key concepts & forms

1. Separable equations: dy/dx = g(x) h(y). Separate and integrate.
2. Homogeneous equations: dy/dx = F(y/x). Use substitution y = vx.
3. Exact equations: M(x,y)dx + N(x,y)dy = 0, check ∂M/∂y = ∂N/∂x.
4. Linear (1st order): dy/dx + P(x) y = Q(x), integrating factor: μ(x)=e^{∫P(x)dx}.
5. Bernoulli: dy/dx + P y = Q y^n → transform with z = y^{1-n}.
6. Cauchy–Euler: x^2 y'' + ax y' + by = 0, try y=x^m.

Theorems & Proofs (compact)

Theorem 1 — Integrating factor for linear first-order

Statement: For dy/dx + P(x)y = Q(x) the integrating factor is μ(x)=e^{∫P(x)dx}. Multiplying transforms LHS into d/dx(μ y).

Proof (brief): Multiply both sides by μ, choose μ so that d(μ)/dx = μ P. This gives d(μ y)/dx = μ Q. Integrate both sides to get μ y = ∫ μ Q dx + C.

SVG Diagrams (unit circle & triangle method)

Formula summary table

परिचय — सारांश

ही नोट्स Differential Equations चा संपूर्ण अध्याय कव्ह करतात — सिद्धांत, साक्षात्कार, 8 सोपे केल्येली उदाहरणे, सराव प्रश्न आणि SVG आकृत्या. परीक्षेच्या दृष्टीने उपयुक्त (SSC, Railway, Police Bharti, NEET).

महत्वाचे तत्त्वे

1. Separable: dy/dx = g(x) h(y). विभाजित करुन समाकलन करा.
2. Homogeneous: dy/dx = F(y/x). y=vx वापर.
3. Exact: Mdx + Ndy = 0, तपासा ∂M/∂y = ∂N/∂x.
4. Linear 1st: dy/dx + P(x)y = Q(x), Integrating factor: μ=e^{∫P(x)dx}.

साधी आठवण

Solved Examples — Step by step (8)

Example 1 — Separable (exponential growth)

Problem: Solve dy/dx = k y where k is constant.

Solution: Separate variables: dy/y = k dx. Integrate both sides: ln|y| = kx + C. So y = C e^{kx}.

Example 2 — Separable

Problem: Solve dy/dx = y cos x.

Solution: dy/y = cos x dx. Integrate: ln|y| = sin x + C. Hence y = C e^{sin x}.

Example 3 — Linear first order (integrating factor)

Problem: Solve dy/dx + y tan x = sin x.

Solution:

1. Standard form: dy/dx + P(x) y = Q(x) with P(x)=tan x, Q(x)=sin x.
2. Integrating factor: μ(x)=e^{∫P dx}=e^{∫tan x dx}=e^{-ln|cos x|}=1/|cos x|. Take μ=1/cos x (assume cos x >0 region).
3. Multiply: (1/cos x) dy/dx + (tan x / cos x) y = tan x. Left side is d/dx (y/cos x).
4. Integrate: y/cos x = ∫ tan x dx = -ln|cos x| + C.
5. Thus y = cos x (-ln|cos x| + C).

Example 4 — Exact equation

Problem: Solve (2xy + 3x^2) dx + (x^2 + 4y) dy = 0.

Solution:

1. Check exactness: ∂M/∂y = 2x, ∂N/∂x = 2x — exact.
2. Find potential Φ(x,y): Φ_x = M = 2xy + 3x^2. Integrate w.r.t x: Φ = x^2 y + x^3 + h(y).
3. Differentiate w.r.t y: Φ_y = x^2 + h'(y) = N = x^2 + 4y. So h'(y)=4y → h=2y^2.
4. Solution: x^2 y + x^3 + 2y^2 = C.

Example 5 — Homogeneous equation

Problem: Solve dy/dx = (x + y)/(x - y).

Solution:

1. Substitute y = v x → dy/dx = v + x dv/dx.
2. RHS becomes (1 + v)/(1 - v). So v + x dv/dx = (1+v)/(1-v).
3. Simplify: x dv/dx = (1+v)/(1-v) - v = (1+v - v(1-v))/(1-v) = (1+v^2)/(1-v).
4. Separate variables: ((1-v)/(1+v^2)) dv = dx/x. Integrate both sides.
5. Left integral: ∫(1/(1+v^2) dv) - ∫(v/(1+v^2) dv) = arctan v - (1/2) ln(1+v^2).
6. So arctan(v) - 1/2 ln(1+v^2) = ln|x| + C. Replace v=y/x for implicit solution.

Example 6 — Second order linear with constant coefficients

Problem: Solve y'' - 3y' + 2y = 0.

Solution: Characteristic equation r^2 - 3r + 2 = 0 → roots r=1,2. General solution: y = C_1 e^{x} + C_2 e^{2x}.

Example 7 — Cauchy–Euler equation

Problem: Solve x^2 y'' - 3x y' + 4 y = 0 for x > 0.

Solution: Try y = x^m. Then characteristic: m(m-1) - 3m + 4 = 0 → m^2 - 4m +4 = 0 → (m-2)^2 = 0. Double root m=2. General solution: y = (C_1 + C_2 ln x) x^2.

Example 8 — Bernoulli equation

Problem: Solve dy/dx + y = y^2 e^{x}.

Solution:

1. Rewrite: dy/dx + P y = Q y^n with P=1, Q=e^{x}, n=2.
2. Put in Bernoulli form: divide by y^2: y^{-2} dy/dx + y^{-1} = e^{x}. Note d/dx(y^{-1}) = -y^{-2} dy/dx.
3. Let z = y^{-1}. Then dz/dx - z = - e^{x}.
4. This is linear in z. Integrating factor μ=e^{-∫1 dx}=e^{-x}. Multiply and integrate: d/dx(z e^{-x}) = -1 → z e^{-x} = -x + C.
5. So z = e^{x}(-x + C) and y = 1/z = e^{-x}/(C - x).

Practice Questions

1. Solve dy/dx = (2x)/(1+y^2).
2. Solve dy/dx + 2y = x with y(0)=1.
3. Check if (y + x)dx + (x - y)dy = 0 is exact. If not, find an integrating factor if possible.
4. Solve dy/dx = (x^2 + y^2)/(xy) (Hint: homogeneous).
5. Solve x^2 y'' - x y' + y = 0 (Cauchy–Euler).
6. Find general solution of y'' + 4y = 0.

Answer Key (brief)

1. ∫(1+y^2) dy = ∫ 2x dx ⇒ y + y^3/3 = x^2 + C (implicit).
2. Integrating factor μ=e^{2x}. Solve: y e^{2x} = ∫ x e^{2x} dx + C → compute gives y = (x/2 - 1/4) + Ce^{-2x} adjust to meet initial condition → y = (x/2 - 1/4) + (5/4)e^{-2x} (verify arithmetic in full solution on page).
3. Not exact: compute partials; integrating factor μ(x)=1/(x^2 + y^2) may work (exercise to verify) — see full worked steps in extended notes.
4. Homogeneous: set y=vx, solve for v(x) and integrate — implicit solution.
5. Try y=x^m, gives equation in m; solve quadratic for m and write general solution.
6. Characteristic r^2 + 4 = 0 → r=±2i → y = C_1 cos 2x + C_2 sin 2x.

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