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State Board · Class 12 · Mathematics Part 2

Lesson 6: Differential Equations

4 concepts · 32 practice questions with answers · Free to read

Order, degree and formation of a differential equation

From daily life

The cooling of tea is described by an equation linking its temperature with the rate at which the temperature falls; that is a differential equation.

Simple definition

A differential equation contains derivatives. Its order is the highest derivative present; its degree is the power of that highest derivative once the equation is free of radicals and fractions in derivatives. A general solution has as many arbitrary constants as the order.

Rule 1: Formation — Differentiate the family of curves as many times as it has arbitrary constants, then eliminate the constants.

How is the formula obtained? : y = Ae^x: y′ = Ae^x = y, so the family satisfies dy/dx = y (order 1, degree 1). y = mx + c (two constants): y′ = m, y″ = 0, so d²y/dx² = 0.

  • Easy : dy/dx + y = 0: order 1, degree 1.
  • Medium : (y″)² + (y′)³ = x: order 2, degree 2.
  • Exam-type : The family y = Ax² + B: y′ = 2Ax, y″ = 2A; eliminating gives x y″ = y′... (one constant removed by y′/x = y″).
Remember

Order counts the highest derivative; degree its power.

Common mistake

Counting the power of y or x as the degree.

Shortcut

Number of arbitrary constants = order of the equation. Memory clue : "Constants equal order."

CET trap

Degree is defined only when the equation is a polynomial in the derivatives.

CET-style example : Order and degree of (y″)² + (y′)³ = x are (A) 2, 2 (B) 2, 3 (C) 3, 2 (D) 1, 2

Answer : (A). Time-saving method : Highest derivative is y″, its power is 2.

Spot the difference

TermMeaning
Orderhighest derivative
Degreepower of the highest derivative

Practice questions

Easy
  1. 1Order of dy/dx + y = 0?
  2. 2Degree of dy/dx + y = 0?
  3. 3Order of d²y/dx² + y = 0?
Medium
  1. 4Order and degree of (y″)² + (y′)³ = x?
  2. 5Differential equation of y = Ae^x.
  3. 6Differential equation of y = mx (m arbitrary).
Challenging
  1. 7Differential equation of y = mx + c.
  2. 8Differential equation of y = A cos x + B sin x.
Show answers (try them yourself first)
  1. 11.
  2. 21.
  3. 32.
  4. 4Order 2, degree 2.
  5. 5dy/dx = y.
  6. 6y = x dy/dx.
  7. 7d²y/dx² = 0.
  8. 8y″ + y = 0.

Variable-separable equations

From daily life

In a population growing at a rate proportional to its size, the variable P and the time t can be moved to separate sides before integrating.

Simple definition

If dy/dx = f(x)·g(y), write dy/g(y) = f(x) dx and integrate both sides. A particular solution fixes the constant using a given condition.

Rule 1: Separate — Collect all y with dy and all x with dx.

Rule 2: Integrate — Integrate both sides and add one constant.

How is the formula obtained? : dy/dx = 2xy: dy/y = 2x dx ⇒ log y = x² + C ⇒ y = Ke^{x²}.

  • Easy : dy/dx = 3x² with y(0) = 1: y = x³ + C, C = 1 ⇒ y = x³ + 1.
  • Medium : dy/dx = y/x: log y = log x + C ⇒ y = kx.
  • Exam-type : dy/dx = 2xy: y = Ke^{x²}.
Remember

Only one arbitrary constant is needed after integrating both sides.

Common mistake

Dividing by y and forgetting y = 0 as a special solution.

Shortcut

Rewrite as ∫dy/g(y) = ∫f(x)dx. Memory clue : "y with dy, x with dx."

CET trap

Write the constant as log C when logs appear, to simplify.

CET-style example : Solution of dy/dx = y/x is (A) y = kx (B) y = k/x (C) y = x + k (D) y = kx²

Answer : (A). Time-saving method : Separate: dy/y = dx/x.

Spot the difference

SolutionConstants
Generalarbitrary constant(s)
Particularconstants fixed by conditions

Practice questions

Easy
  1. 1Solve dy/dx = 3x² with y(0) = 1.
  2. 2Solve dy/dx = y.
  3. 3Solve dy/dx = y/x.
Medium
  1. 4Solve dy/dx = 2xy.
  2. 5Solve dy/dx = x/y.
  3. 6Solve dy/dx = e^{x − y}.
Challenging
  1. 7Solve dy/dx = (1 + y²) with y(0) = 0.
  2. 8Solve dy/dx = y with y(0) = 3.
Show answers (try them yourself first)
  1. 1y = x³ + 1.
  2. 2y = Ke^x.
  3. 3y = kx.
  4. 4y = Ke^{x²}.
  5. 5y dy = x dx ⇒ y² = x² + c.
  6. 6e^y dy = e^x dx ⇒ e^y = e^x + C.
  7. 7tan⁻¹y = x ⇒ y = tan x.
  8. 8y = 3e^x.

Homogeneous differential equations

From daily life

When both the numerator and denominator scale together, as in a recipe doubled for a bigger party, a ratio y/x captures the whole behaviour.

Simple definition

dy/dx = f(y/x) is homogeneous. Put y = vx, so dy/dx = v + x dv/dx, which gives a variable-separable equation in v and x. Finally put v = y/x.

Rule 1: Test — Replace x by kx and y by ky; if the ratio is unchanged the equation is homogeneous.

Rule 2: Substitute — y = vx ⇒ v + x dv/dx = f(v).

How is the formula obtained? : dy/dx = (x + y)/x: v + x dv/dx = 1 + v ⇒ x dv/dx = 1 ⇒ dv = dx/x ⇒ v = log x + C ⇒ y = x(log x + C).

  • Easy : dy/dx = y/x is homogeneous (v = y/x).
  • Medium : dy/dx = (x + y)/x gives y = x(log x + C).
  • Exam-type : dy/dx = (x² + y²)/(2xy): v + x v′ = (1 + v²)/(2v) ⇒ x v′ = (1 − v²)/(2v) ⇒ ∫2v/(1 − v²) dv = log x + c ⇒ x² − y² = Cx.
Remember

Always return to y/x at the end.

Common mistake

Using v = x/y without adjusting the equation.

Shortcut

Divide numerator and denominator by x to see f(y/x). Memory clue : "y = vx, then separate."

CET trap

x v′ is not a derivative of v alone; remember dv/dx multiplies x.

CET-style example : dy/dx = (x + y)/x has solution (A) y = x(log x + C) (B) y = x + C (C) y = C/x (D) y = x²

Answer : (A). Time-saving method : v + x v′ = 1 + v.

Spot the difference

TypeSubstitution
Homogeneousy = vx
Variable separablenone needed

Practice questions

Easy
  1. 1Is dy/dx = (x + y)/x homogeneous?
  2. 2Substitution for homogeneous equations?
  3. 3Then dy/dx = ?
Medium
  1. 4Solve dy/dx = (x + y)/x.
  2. 5Solve dy/dx = y/x + 1 given y(1) = 0.
  3. 6Solve x dy/dx = y + x.
Challenging
  1. 7Solve dy/dx = (x² + y²)/(2xy).
  2. 8Check y = x log x against dy/dx = y/x + 1.
Show answers (try them yourself first)
  1. 1Yes; it equals 1 + y/x.
  2. 2y = vx.
  3. 3v + x dv/dx.
  4. 4y = x(log x + C).
  5. 5y = x log x.
  6. 6y = x(log x + C).
  7. 7x² − y² = Cx.
  8. 8y′ = log x + 1 = y/x + 1 ✓.

Linear differential equations of the first order

From daily life

A tank filling while draining at a rate linked to its own level is a linear equation: the rate is a sum of a term in y and a term in x.

Simple definition

dy/dx + P y = Q (P, Q functions of x) is linear. The integrating factor is I.F. = e^{∫P dx}. The solution is y·(I.F.) = ∫Q·(I.F.) dx + C.

Rule 1: Standard form — Write the equation as dy/dx + P y = Q before finding I.F.

Rule 2: Solve — y × I.F. = ∫ Q × I.F. dx + C.

How is the formula obtained? : dy/dx + 2y = e^{−x}: P = 2, I.F. = e^{2x}. y e^{2x} = ∫e^{x} dx = e^{x} + C ⇒ y = e^{−x} + C e^{−2x}.

  • Easy : dy/dx + y = 0: I.F. = e^x; y e^x = C ⇒ y = Ce^{−x}.
  • Medium : dy/dx + y/x = x²: I.F. = x; xy = x⁴/4 + C ⇒ y = x³/4 + C/x.
  • Exam-type : dy/dx + 2y = e^{−x}: y = e^{−x} + Ce^{−2x}.
Remember

The integrating factor makes the left side the derivative of y × I.F.

Common mistake

Not writing the equation with coefficient 1 on dy/dx first.

Shortcut

e^{log f} = f simplifies the I.F. for P = 1/x (gives x). Memory clue : "I.F. is e to the integral of P."

CET trap

Do not add a constant when finding the I.F.

CET-style example : I.F. of dy/dx + y/x = x² is (A) x (B) 1/x (C) e^x (D) log x

Answer : (A). e^{log x} = x. Time-saving method : Use e^{∫(1/x)dx}.

Spot the difference

PartMeaning
Pcoefficient of y
Qright-hand side
I.F.e^{∫P dx}

Practice questions

Easy
  1. 1I.F. of dy/dx + y = x?
  2. 2I.F. of dy/dx + 2y = e^{−x}?
  3. 3I.F. of dy/dx + y/x = x²?
Medium
  1. 4Solve dy/dx + y = 0.
  2. 5Solve dy/dx + y/x = x².
  3. 6Solve dy/dx + 2y = e^{−x}.
Challenging
  1. 7Solve dy/dx + y = x.
  2. 8Solve dy/dx + y/x = 1 with y(1) = 2.
Show answers (try them yourself first)
  1. 1e^x.
  2. 2e^{2x}.
  3. 3x.
  4. 4y = Ce^{−x}.
  5. 5y = x³/4 + C/x.
  6. 6y = e^{−x} + Ce^{−2x}.
  7. 7y = x − 1 + Ce^{−x}.
  8. 8xy = x²/2 + C; 2 = ½ + C ⇒ y = x/2 + 3/(2x).