Differential Equations: CBSE Class 12 Maths

Fundamental Concepts: Order and Degree

Differential Equation

An equation involving an independent variable (like x), a dependent variable (like y), and the derivatives of the dependent variable with respect to the independent variable (e.g., dy/dx, d²y/dx²).

Order

The order of a differential equation is the order of the highest derivative appearing in the equation. For example, in d²y/dx² + 3(dy/dx) + 2y = 0, the highest derivative is d²y/dx², so the order is 2.

Degree

The degree of a differential equation is the highest power (exponent) of the highest order derivative, provided the equation is a polynomial in its derivatives. For (d³y/dx³)² + dy/dx = 0, the order is 3 and the degree is 2. However, for sin(dy/dx) + y = 0, the degree is not defined because it's not a polynomial in dy/dx.

Methods for Solving First-Order, First-Degree Differential Equations

For your Class 12 syllabus, we primarily focus on solving first-order, first-degree differential equations. There are three main methods you need to master.

1. Variable Separable Method: This is the simplest technique. The goal is to rearrange the equation so that all terms involving the variable y and dy are on one side, and all terms involving x and dx are on the other. Once separated, you can integrate both sides to find the general solution. The form looks like f(y) dy = g(x) dx.

1. Homogeneous Differential Equations: An equation of the form dy/dx = F(x, y) is homogeneous if F(x, y) is a homogeneous function of degree zero. A practical test is to see if you can write the function as g(y/x) or h(x/y). To solve these, we use the substitution y = vx. This transforms the equation into a variable separable form in terms of v and x, which we can then solve.

1. Linear Differential Equations: These equations are of the standard form dy/dx + Py = Q, where P and Q are functions of x (or constants). These are solved using a special function called the Integrating Factor (I.F.), which is calculated as I.F. = e^(∫P dx). Multiplying the entire equation by the I.F. magically turns the left-hand side into the derivative of y (I.F.). This allows us to integrate and find the solution easily. The final solution is given by the formula: y (I.F.) = ∫(Q * I.F.) dx + C.

Step-by-Step: Solving a Linear Differential Equation

1. Step 1: Standard Form — Rearrange the given differential equation into the standard linear form: dy/dx + Py = Q. Make sure the coefficient of dy/dx is 1.
2. Step 2: Identify P and Q — Once in standard form, identify the functions P and Q. Remember, both P and Q must be functions of x only, or constants.
3. Step 3: Calculate the Integrating Factor (I.F.) — Use the formula I.F. = e^(∫P dx) to find the Integrating Factor. Carefully evaluate the integral of P.
4. Step 4: Apply the Solution Formula — Substitute the I.F. and Q into the general solution formula: y (I.F.) = ∫(Q I.F.) dx + C.
5. Step 5: Integrate and Solve — Evaluate the integral on the right-hand side. This will often require integration by parts or other techniques. Finally, simplify the equation to express y in terms of x.

Worked Examples of Differential Equations

• Example 1 (Variable Separable): Solve the differential equation dy/dx = (1+x²)(1+y²). Step 1: Separate the variables. Rearrange the equation to get y-terms with dy and x-terms with dx. dy / (1+y²) = (1+x²) dx Step 2: Integrate both sides. ∫ dy / (1+y²) = ∫ (1+x²) dx Step 3: Perform the integration. We know ∫ 1/(1+u²) du = tan⁻¹(u). So, tan⁻¹(y) = ∫ 1 dx + ∫ x² dx tan⁻¹(y) = x + (x³/3) + C Final Answer: The general solution is tan⁻¹(y) = x + x³/3 + C.
• Example 2 (Homogeneous): Solve x dy - y dx = √(x² + y²) dx. Step 1: Rearrange and check for homogeneity. dy/dx = (y + √(x² + y²)) / x dy/dx = y/x + √(1 + (y/x)²). This is in the form g(y/x), so it's homogeneous. Step 2: Substitute y = vx. This means dy/dx = v + x(dv/dx). Substituting into the equation: v + x(dv/dx) = v + √(1 + v²) Step 3: Simplify and separate variables. x(dv/dx) = √(1 + v²) dv / √(1 + v²) = dx / x Step 4: Integrate both sides. ∫ dv / √(1 + v²) = ∫ dx / x Using the standard integral formula, we get: log|v + √(1 + v²)| = log|x| + log|C| (using log C for convenience) log|v + √(1 + v²)| = log|Cx| v + √(1 + v²) = Cx Step 5: Substitute back v = y/x. (y/x) + √(1 + (y/x)²) = Cx (y/x) + (√(x² + y²))/x = Cx Final Answer: The general solution is y + √(x² + y²) = Cx².

CBSE Exam Traps and Common Mistakes

Students often lose marks due to simple oversights. Here are the top three traps to avoid:

• Forgetting the Constant of Integration +C: This is the most common mistake. Every indefinite integration must include the constant C. In problems where a particular solution is required (i.e., initial conditions like y(0)=1 are given), you use these conditions to find the value of C. Forgetting it means your general solution is incorrect from the start.
• Incorrectly Identifying the Equation Type: Don't rush! Before solving, take a moment to classify the equation. Is it variable separable? Homogeneous? Linear? Trying to solve a linear equation with the variable separable method will lead nowhere. Check for homogeneity using the F(λx, λy) test or by trying to express it as g(y/x). Check for the linear form dy/dx + Py = Q.
• Mistakes in Integration: Differential equations bring together all your integration skills. Be extra careful with the integration of P for the Integrating Factor (e^(∫P dx)) and the final integration of ∫(Q * I.F.) dx. This step often requires integration by parts, and a small error can derail the entire solution.

Practice Questions with Solutions

• Q: Find the sum of the order and degree of the differential equation: (d²y/dx²)³ + (dy/dx)⁴ + sin(x) = 0. A: Step 1: Identify the highest order derivative. The derivatives present are d²y/dx² and dy/dx. The highest order is 2. Step 2: Determine the degree. The equation is a polynomial in its derivatives. The power of the highest order derivative (d²y/dx²) is 3. So, the degree is 3. Step 3: Calculate the sum. Sum = Order + Degree = 2 + 3 = 5. Final answer: 5
• Q: Find the general solution of the differential equation: dy/dx = e^(x-y). A: Step 1: Recognize the type of equation. The equation can be written as dy/dx = e^x * e^(-y). This is a variable separable equation. Step 2: Separate the variables. Move all y-terms to the left and x-terms to the right. e^y dy = e^x dx Step 3: Integrate both sides. ∫e^y dy = ∫e^x dx Step 4: Perform the integration. e^y = e^x + C Final answer: The general solution is e^y = e^x + C.
• Q: Solve the differential equation: (x² + xy) dy = (x² + y²) dx. A: Step 1: Rearrange into dy/dx form and check for homogeneity. dy/dx = (x² + y²) / (x² + xy). Dividing numerator and denominator by x² gives dy/dx = (1 + (y/x)²) / (1 + y/x). This is a homogeneous function of y/x. Step 2: Substitute y = vx, which implies dy/dx = v + x(dv/dx). v + x(dv/dx) = (1 + v²) / (1 + v) Step 3: Simplify and separate variables. x(dv/dx) = (1 + v²) / (1 + v) - v = (1 + v² - v - v²) / (1 + v) = (1 - v) / (1 + v) ( (1 + v) / (1 - v) ) dv = dx / x ( (- (v - 1 - 2)) / (v - 1) ) dv = -dx / x => ( -1 - 2/(v-1) ) dv = -dx/x ( 1 + 2/(v-1) ) dv = dx/x Step 4: Integrate both sides. ∫(1 + 2/(v-1)) dv = ∫(1/x) dx v + 2 log|v-1| = log|x| + C Step 5: Substitute back v = y/x. y/x + 2 log|(y/x) - 1| = log|x| + C Final answer: y/x + 2 log|(y-x)/x| = log|x| + C, which can be further simplified.
• Q: Find the particular solution of dy/dx + 2y tan(x) = sin(x), given that y=0 when x=π/3. A: Step 1: Identify the type. The equation is in the linear form dy/dx + Py = Q, with P = 2 tan(x) and Q = sin(x). Step 2: Calculate the Integrating Factor (I.F.). I.F. = e^(∫P dx) = e^(∫2 tan(x) dx) = e^(2 log|sec(x)|) = e^(log(sec²(x))) = sec²(x). Step 3: Apply the general solution formula: y (I.F.) = ∫(Q I.F.) dx + C. y sec²(x) = ∫(sin(x) sec²(x)) dx + C y sec²(x) = ∫(tan(x) sec(x)) dx + C Step 4: Integrate the right side. y sec²(x) = sec(x) + C This is the general solution. Step 5: Find the particular solution using y=0, x=π/3. 0 sec²(π/3) = sec(π/3) + C 0 = 2 + C => C = -2. Step 6: Substitute C back into the general solution. y sec²(x) = sec(x) - 2 Final answer: The particular solution is y = cos(x) - 2cos²(x).

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