This sheet covers the differential equations class 12 formulas you need for quick revision: order and degree, general and particular solutions, and the three standard methods for first order–first degree equations — variables separable, homogeneous, and linear with an integrating factor.
Each formula below is grouped by topic, with the meaning of every symbol, when-to-use guidance, and worked examples using original numbers. This page is part of the Class 12 mathematics formulas collection; browse the maths formulas index for formula sheets of other classes and chapters.
Differential Equations Class 12 Formulas at a Glance
The table below is a quick index to the formulas covered on this page, in the same order as the full list further down.
| Purpose | Formula |
|---|---|
| Order of a differential equation | Order of the highest order derivative present |
| Degree (only when the equation is a polynomial in derivatives) | Highest power of the highest order derivative |
| Shorthand for derivatives | \( y’ = \frac{dy}{dx},\ y” = \frac{d^2y}{dx^2},\ y_n = \frac{d^n y}{dx^n} \) |
| Variables separable form | \( \frac{dy}{dx} = h(y)\,g(x) \) |
| Separate and integrate (when \( h(y) \neq 0 \)) | \( \int \frac{1}{h(y)}\,dy = \int g(x)\,dx \) |
| Test for a homogeneous function of degree \( n \) | \( F(\lambda x, \lambda y) = \lambda^n F(x, y) \) |
| Homogeneous differential equation form | \( \frac{dy}{dx} = g\left( \frac{y}{x} \right) \) |
| Substitution that solves it | \( y = vx,\quad \frac{dy}{dx} = v + x\frac{dv}{dx} \) |
| Reduced separable equation | \( \int \frac{dv}{g(v) – v} = \int \frac{dx}{x} \) |
| First order linear form | \( \frac{dy}{dx} + Py = Q \) |
| Alternative linear form (x as dependent variable) | \( \frac{dx}{dy} + P_1 x = Q_1 \) |
| Integrating factor | \( \text{I.F.} = e^{\int P\,dx} \) |
| General solution of the linear form | \( y\,(\text{I.F.}) = \int (Q \times \text{I.F.})\,dx + C \) |
| Equivalent closed form of the linear solution | \( y = e^{-\int P\,dx}\int\left( Q\,e^{\int P\,dx} \right)dx + C \) |
| Integrating factor and solution for the \( x \)-form | \( \text{I.F.} = e^{\int P_1 dy},\quad x\,(\text{I.F.}) = \int (Q_1 \times \text{I.F.})\,dy + C \) |
All Formulas, Grouped by Topic
Every formula below follows the NCERT Class 12 Mathematics Part II textbook (Rationalised NCERT). You can cross-check any of them in the official NCERT PDF of this chapter (file lemh203.pdf).
Basic Concepts: Order and Degree
A differential equation is an equation involving derivatives of the dependent variable with respect to the independent variable (NCERT, p. 301). The chapter uses this shorthand for derivatives (NCERT, p. 301):
\[ y’ = \frac{dy}{dx}, \qquad y” = \frac{d^2y}{dx^2}, \qquad y”’ = \frac{d^3y}{dx^3}, \qquad y_n = \frac{d^n y}{dx^n} \]
Order is the order of the highest order derivative present in the equation (NCERT, p. 301).
Degree is defined only when the equation is a polynomial in its derivatives; it is then the highest power (positive integral index) of the highest order derivative (NCERT, p. 302). In \( \left( \frac{d^2y}{dx^2} \right)^3 + \left( \frac{dy}{dx} \right)^2 + y = 0 \), for instance, the order is 2 and the degree is 3.
Order and degree (when defined) are always positive integers (NCERT, p. 302).
General and Particular Solutions
A function \( y = \phi(x) \) that satisfies the equation is called a solution, and its graph is the solution curve. The general solution (primitive) contains arbitrary constants — as many as the order of the equation — while the particular solution is free of arbitrary constants and is obtained by giving particular values to those constants (NCERT, p. 305; p. 337).
Differential Equations with Variables Separable
A first order–first degree equation is of the form \( \frac{dy}{dx} = F(x, y) \). It is of variable separable type when \( F(x, y) \) can be expressed as a product of a function of \( x \) and a function of \( y \) (NCERT, p. 307):
\[ \frac{dy}{dx} = h(y)\,g(x) \]
For \( h(y) \neq 0 \), separate the variables and integrate both sides (NCERT, p. 307):
\[ \int \frac{1}{h(y)}\,dy = \int g(x)\,dx \]
The solution is \( H(y) = G(x) + C \), where \( H \) and \( G \) are antiderivatives of \( 1/h(y) \) and \( g(x) \) respectively (NCERT, p. 307).
Homogeneous Differential Equations
A function \( F(x, y) \) is a homogeneous function of degree \( n \) if (NCERT, p. 312):
\[ F(\lambda x, \lambda y) = \lambda^n F(x, y) \quad \text{for any nonzero constant } \lambda \]
Equivalently, \( F(x, y) = x^n\,g\left( \frac{y}{x} \right) \) or \( y^n\,h\left( \frac{x}{y} \right) \) (NCERT, p. 313).
A differential equation \( \frac{dy}{dx} = F(x, y) \) is homogeneous when \( F(x, y) \) is a homogeneous function of degree zero, i.e. when it can be written as \[ \frac{dy}{dx} = g\left( \frac{y}{x} \right) \]
To solve it, substitute \( y = vx \) and use \( \frac{dy}{dx} = v + x\frac{dv}{dx} \) (NCERT, p. 313):
\[ \int \frac{dv}{g(v) – v} = \int \frac{dx}{x} \]
After integrating, replace \( v \) by \( y/x \). If the equation is written as \( \frac{dx}{dy} = h\left( \frac{x}{y} \right) \), substitute \( x = vy \) instead (NCERT, p. 313).
Linear Differential Equations
A first order linear differential equation has the form (NCERT, p. 322):
\[ \frac{dy}{dx} + Py = Q \]
where \( P \) and \( Q \) are constants or functions of \( x \) only. The alternate form, used when \( x \) is treated as the dependent variable, is (NCERT, p. 322):
\[ \frac{dx}{dy} + P_1 x = Q_1 \]
where \( P_1 \) and \( Q_1 \) are constants or functions of \( y \) only. Multiply the first form by the integrating factor (NCERT, p. 323):
\[ \text{I.F.} = e^{\int P\,dx} \]
The factor is chosen so that the left side becomes the derivative of \( y \cdot \text{I.F.} \), which is why the method works. The general solution is (NCERT, p. 323):
\[ y\,(\text{I.F.}) = \int (Q \times \text{I.F.})\,dx + C \]
or, in equivalent closed form, \[ y = e^{-\int P\,dx}\int\left( Q\,e^{\int P\,dx} \right)dx + C \]
For the \( x \)-form: \( \text{I.F.} = e^{\int P_1 dy} \) and \( x\,(\text{I.F.}) = \int (Q_1 \times \text{I.F.})\,dy + C \) (NCERT, p. 323).
What Each Symbol Means
For a pure mathematics chapter there are no physical SI units; the third column states the nature of each quantity instead.
| Symbol | What it means | Unit / nature |
|---|---|---|
| \( x \) | Independent variable of the equation | Real number (dimensionless) |
| \( y \) | Dependent variable, whose derivatives appear in the equation | Real number (dimensionless) |
| \( y’ = \frac{dy}{dx} \) | First derivative of \( y \) with respect to \( x \) | Rate of change of \( y \) per unit \( x \) |
| \( y” = \frac{d^2y}{dx^2} \) | Second derivative of \( y \) with respect to \( x \) | Rate of change of \( y’ \) per unit \( x \) |
| \( y_n = \frac{d^n y}{dx^n} \) | \( n \)-th order derivative of \( y \) | \( n \)-th order rate of change |
| \( P, Q \) | Coefficients in \( \frac{dy}{dx} + Py = Q \); constants or functions of \( x \) only | Dimensionless functions |
| \( P_1, Q_1 \) | Coefficients in \( \frac{dx}{dy} + P_1x = Q_1 \); constants or functions of \( y \) only | Dimensionless functions |
| \( \text{I.F.} = e^{\int P\,dx} \) | Integrating factor that makes the left side an exact derivative | Dimensionless multiplier |
| \( \lambda \) | Nonzero constant used in the homogeneity test \( F(\lambda x, \lambda y) = \lambda^n F(x, y) \) | Real number |
| \( n \) | Degree of a homogeneous function; also the order of a derivative | Non-negative integer / positive integer |
| \( v \) | Substitution variable: \( v = y/x \) (or \( v = x/y \) in the alternate form) | Dimensionless ratio |
| \( g(x), h(y) \) | Factors of the right side in the separable form \( \frac{dy}{dx} = h(y)\,g(x) \) | Functions of one variable |
| \( H(y), G(x) \) | Antiderivatives of \( 1/h(y) \) and \( g(x) \) | Functions of one variable |
| \( C, C_1 \) | Arbitrary constants of integration | Real numbers |
When to Use Each Formula
| Situation | Formula or method to reach for |
|---|---|
| You are asked to find the order of a differential equation | Order = order of the highest derivative present (NCERT, p. 301) |
| You are asked to find the degree | Degree = highest power of the highest order derivative, but only if the equation is a polynomial in its derivatives (NCERT, p. 302) |
| The right side of \( \frac{dy}{dx} = F(x, y) \) splits into \( g(x) \cdot h(y) \) | Variables separable: \( \int \frac{1}{h(y)}\,dy = \int g(x)\,dx \) (NCERT, p. 307) |
| \( \frac{dy}{dx} = F(x, y) \) with \( F(\lambda x, \lambda y) = F(x, y) \), i.e. the right side is \( g(y/x) \) | Homogeneous: substitute \( y = vx \), integrate \( \int \frac{dv}{g(v) – v} = \int \frac{dx}{x} \), then replace \( v \) by \( y/x \) (NCERT, p. 313) |
| The equation is in the form \( \frac{dx}{dy} = h\left( \frac{x}{y} \right) \) | Homogeneous alternate form: substitute \( x = vy \) (NCERT, p. 313) |
| The equation can be written as \( \frac{dy}{dx} + Py = Q \) with \( P, Q \) functions of \( x \) only | Linear: \( \text{I.F.} = e^{\int P\,dx} \), then \( y \cdot \text{I.F.} = \int (Q \times \text{I.F.})\,dx + C \) (NCERT, p. 323) |
| The equation is linear in \( x \) as a function of \( y \), of the form \( \frac{dx}{dy} + P_1x = Q_1 \) | Linear alternate form: \( \text{I.F.} = e^{\int P_1 dy} \), then \( x \cdot \text{I.F.} = \int (Q_1 \times \text{I.F.})\,dy + C \) (NCERT, p. 323) |
| A condition such as \( y(x_0) = y_0 \) is given | Find the general solution first, substitute the condition to fix \( C \), then write the particular solution (NCERT, p. 305) |
Worked Examples
Example 1: Finding Order and Degree
- Step 1: In \( \left( \frac{d^2y}{dx^2} \right)^3 + 2\left( \frac{dy}{dx} \right)^4 + 7y = 0 \), the highest order derivative is \( \frac{d^2y}{dx^2} \), so the order is 2.
- Step 2: The equation is a polynomial in derivatives (every derivative appears with a positive integer power), so the degree is defined.
The power of the highest order derivative is 3.
Final answer: Order = 2, degree = 3.
Check the not-defined case: in \( \frac{d^3y}{dx^3} + \sin\left( \frac{dy}{dx} \right) = 0 \), the order is 3 but the degree is not defined, because \( \sin(dy/dx) \) is not a polynomial in the derivative (NCERT, p. 302).
Example 2: Variables Separable with a Given Condition
Step 1: The equation \( \frac{dy}{dx} = \frac{x}{y^2} \) has the separable form \( \frac{dy}{dx} = g(x)\,h(y) \).
Since \( h(y) = \frac{1}{y^2} \neq 0 \), separate the variables: \( y^2\,dy = x\,dx \).
Step 2: Integrate both sides: \( \int y^2\,dy = \int x\,dx \), so \( \frac{y^3}{3} = \frac{x^2}{2} + C \), i.e.
\( y^3 = \frac{3}{2}x^2 + 3C \).
Step 3: The condition \( y(0) = 2 \) fixes the constant: \( 8 = 0 + 3C \), so \( C = \frac{8}{3} \).
\[ y^3 = \frac{3}{2}x^2 + 8 \]
Final answer: \( y = \sqrt[3]{\frac{3}{2}x^2 + 8} \). Check: at \( x = 0 \), \( y = \sqrt[3]{8} = 2 \).
Example 3: Linear Equation Using the Integrating Factor
- Step 1: \( \frac{dy}{dx} + y = x \) is already in the linear form \( \frac{dy}{dx} + Py = Q \) with \( P = 1 \), \( Q = x \).
- Step 2: Integrating factor: \( \text{I.F.} = e^{\int P\,dx} = e^{\int 1\,dx} = e^x \).
- Step 3: Apply \( y\,(\text{I.F.}) = \int (Q \times \text{I.F.})\,dx + C \): \( y\,e^x = \int x\,e^x\,dx + C = (x – 1)e^x + C \).
- Step 4: The general solution is \( y = x – 1 + Ce^{-x} \).
The condition \( y(0) = 1 \) gives \( 1 = -1 + C \), so \( C = 2 \).
Final answer: \( y = x – 1 + 2e^{-x} \). Check: \( y’ + y = (1 – 2e^{-x}) + (x – 1 + 2e^{-x}) = x \).
Homogeneous in one line: For \( \frac{dy}{dx} = \frac{x^2 + 3y^2}{2xy} \), rewrite the right side as \( g\left( \frac{y}{x} \right) = \frac{1 + 3(y/x)^2}{2(y/x)} \), substitute \( y = vx \) with \( \frac{dy}{dx} = v + x\frac{dv}{dx} \), separate to \( \frac{2v}{1+v^2}\,dv = \frac{dx}{x} \), integrate and replace \( v \) by \( y/x \) — the solution is \( x^2 + y^2 = Cx^3 \).
Common Mistakes to Avoid
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Writing the degree of \( \left( \frac{d^2y}{dx^2} \right)^3 + \sin\left( \frac{dy}{dx} \right) = 0 \) as 3 or 1 | Degree is defined only when the equation is a polynomial in its derivatives; \( \sin(dy/dx) \) violates that, so the degree is not defined | Scan every derivative — if one sits inside \( \sin \), \( \cos \), \( \log \) or an exponent, write “degree not defined” |
| Confusing order with degree in \( \left( \frac{d^2y}{dx^2} \right)^3 + y’ = 0 \) and writing order 3 | Order is the order of the highest derivative (2 here); degree is the power of that derivative (3) | Name the highest derivative first, then read its exponent — order first, degree second |
| Reading off \( P = 2 \) directly from \( x\frac{dy}{dx} + 2y = x^2 \) | Divide every term by \( x \) first: \( \frac{dy}{dx} + \frac{2}{x}y = x \), so \( P = \frac{2}{x} \) and \( Q = x \) | The coefficient of \( dy/dx \) must be 1 before you identify \( P \) and \( Q \) |
| Writing \( \text{I.F.} = e^{\int P_1 dx} \) for \( \frac{dx}{dy} + P_1x = Q_1 \) | For the \( x \)-form, integrate with respect to \( y \): \( \text{I.F.} = e^{\int P_1 dy} \) | Integrate \( P_1 \) with respect to the same variable that appears in the derivative — \( dy \), not \( dx \) |
| Substituting \( y = vx \) but forgetting that \( \frac{dy}{dx} = v + x\frac{dv}{dx} \), or leaving \( v \) in the final answer | Differentiate the substitution along with it; after integrating, replace \( v \) by \( y/x \) and simplify | The final answer must contain only \( x \), \( y \) and constants — no \( v \) should survive |
| Applying \( y = vx \) to an equation that is not homogeneous of degree zero | First test \( F(\lambda x, \lambda y) = \lambda^0 F(x, y) \), or check that dividing numerator and denominator by the highest power of \( x \) leaves only ratios \( y/x \) | If the right side cannot be written as \( g(y/x) \), the substitution will not work — use another method |
Frequently Asked Questions
When is the degree of a differential equation not defined?
Whenever the equation is not a polynomial in its derivatives. For example, \( \frac{d^3y}{dx^3} + \sin\left( \frac{dy}{dx} \right) = 0 \) has order 3 but its degree is not defined, because \( dy/dx \) appears inside \( \sin \). The same applies when a derivative appears inside \( e^{y’} \) or \( \log(y’) \) (NCERT, p. 302).
How many arbitrary constants are in a general solution?
A general solution of an \( n \)-th order differential equation contains exactly \( n \) arbitrary constants — the number equals the order. A particular solution is free of arbitrary constants, because the given conditions have fixed their values (NCERT, p. 337).
When do I use y = vx and when x = vy?
Use \( y = vx \) when the equation is written as \( \frac{dy}{dx} = g\left( \frac{y}{x} \right) \), and use \( x = vy \) when it is written as \( \frac{dx}{dy} = h\left( \frac{x}{y} \right) \). In either case, replace \( v \) by the original ratio at the end so the answer contains only \( x \) and \( y \).
Why does the integrating factor work?
Multiplying \( \frac{dy}{dx} + Py = Q \) by \( e^{\int P\,dx} \) converts the left side into the derivative of \( y\,e^{\int P\,dx} \). Both sides can then be integrated directly, which is exactly why that factor is chosen (NCERT, p. 323).
Reference: NCERT Class 12 Mathematics textbook, chapter Differential Equations.
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