NCERT Class 12 Maths: Differential Equations - Exercise 9.4

Understanding the Variable Separable Method

A differential equation is an equation that involves an unknown function and its derivatives. In Class 12, we primarily deal with first-order, first-degree differential equations, which involve only the first derivative, usually written as dy/dx or dx/dy. The Variable Separable Method is one of the simplest and most common techniques to solve such differential equations. A differential equation is said to be of the variable separable type if it can be written in the form f(y) dy = g(x) dx. This means that all terms involving the variable y and dy can be grouped on one side of the equation, and all terms involving the variable x and dx can be grouped on the other side. The core idea is to literally 'separate' the variables so that each side of the equation contains only one type of variable and its corresponding differential. Once separated, integrating both sides independently with respect to their respective variables allows us to find the general solution of the differential equation. This general solution will contain an arbitrary constant, 'C'. If initial conditions (specific values of x and y) are provided, we can use them to find the value of 'C' and obtain a 'particular solution'.

Steps to Solve Differential Equations Using Variable Separable Method

1. Step 1: Identify and Rearrange — Examine the given differential equation. Try to rearrange it into the form dy/dx = f(x)g(y) or dx/dy = f(x)g(y). This involves algebraic manipulations like multiplication, division, or factorization to isolate terms.
2. Step 2: Separate the Variables — Once in the form dy/dx = f(x)g(y), move all terms involving y (including dy) to one side and all terms involving x (including dx) to the other side. This results in the form (1/g(y)) dy = f(x) dx (or f(y) dy = g(x) dx if rearranged differently).
3. Step 3: Integrate Both Sides — Integrate both sides of the separated equation with respect to their respective variables. Remember to include a single constant of integration, 'C', on one side (usually the right-hand side is convenient) after integration. For example, ∫(1/g(y)) dy = ∫f(x) dx + C.
4. Step 4: Solve for the Unknown Function (if possible) — After integration, you will have an equation relating y and x and the constant C. This is the general solution. If possible and required, express y explicitly as a function of x. If initial conditions are given, substitute them into the general solution to find the value of C and obtain the particular solution.

Worked Examples: Applying the Variable Separable Method

• Example 1: Solve the differential equation dy/dx = (1+y^2)/(1+x^2). Step 1: Identify and Rearrange The equation is already in a suitable form, dy/dx = g(y)f(x), where g(y) = 1+y^2 and f(x) = 1/(1+x^2). Step 2: Separate the Variables Move (1+y^2) to the left side and dx to the right side: dy / (1+y^2) = dx / (1+x^2) Step 3: Integrate Both Sides Integrate both sides: ∫ dy / (1+y^2) = ∫ dx / (1+x^2) Recall that ∫ 1/(a^2+x^2) dx = (1/a)tan⁻¹(x/a). Applying this formula: tan⁻¹(y) = tan⁻¹(x) + C Step 4: Solve for y (General Solution) The general solution is tan⁻¹(y) = tan⁻¹(x) + C. Alternatively, y = tan(tan⁻¹(x) + C).
• Example 2: Find the particular solution of (e^x + e^-x)dy - (e^x - e^-x)dx = 0, given that y=2 when x=0. Step 1: Identify and Rearrange First, rearrange the equation to isolate dy and dx terms: (e^x + e^-x)dy = (e^x - e^-x)dx Step 2: Separate the Variables Divide both sides by (e^x + e^-x) and dx to get the dy/dx form, then separate: dy = ( (e^x - e^-x) / (e^x + e^-x) ) dx Variables are now separated: f(y) dy = g(x) dx where f(y) = 1. Step 3: Integrate Both Sides ∫ dy = ∫ ( (e^x - e^-x) / (e^x + e^-x) ) dx For the right-hand side integral, let u = e^x + e^-x. Then du = (e^x - e^-x)dx. So, ∫ du/u = ln|u|. Integrating both sides: y = ln|e^x + e^-x| + C This is the general solution. Step 4: Find the Particular Solution Use the given initial condition: y=2 when x=0. 2 = ln|e^0 + e^-0| + C 2 = ln|1 + 1| + C 2 = ln(2) + C C = 2 - ln(2) Substitute the value of C back into the general solution: y = ln|e^x + e^-x| + (2 - ln(2)) This is the particular solution.
• Example 3: Solve the differential equation dy/dx = (x+1)/(2-y). Step 1: Identify and Rearrange The equation is in the form dy/dx = f(x)/g(y). Step 2: Separate the Variables Multiply both sides by (2-y) and dx: (2-y) dy = (x+1) dx Step 3: Integrate Both Sides ∫ (2-y) dy = ∫ (x+1) dx Integrate term by term: 2y - (y^2)/2 = (x^2)/2 + x + C Step 4: Express General Solution (Implicit Form) To eliminate fractions, multiply the entire equation by 2: 4y - y^2 = x^2 + 2x + 2C Let 2C = K (another arbitrary constant). 4y - y^2 = x^2 + 2x + K Rearranging to group terms: x^2 + y^2 + 2x - 4y + K = 0 This is the general solution in implicit form.

Exam Tips for Differential Equations (Ex 9.4)

When solving problems from Exercise 9.4, keep these crucial tips in mind to avoid common errors and score well:

1. Don't Forget the Constant of Integration (C): This is one of the most frequent mistakes. Every indefinite integral must have a constant C. Place it on only one side of the equation, usually the side involving x. If you're asked for a particular solution, remember to use the given initial conditions to find the value of C.
2. Algebraic Manipulation is Key: Before integrating, ensure you have correctly separated the variables. Sometimes, a complex equation might require factorization, cross-multiplication, or rearrangement to bring it into the f(y) dy = g(x) dx form. Take your time with these steps.
3. Master Integration Formulas: The heart of this method lies in correct integration. Be thoroughly familiar with standard integral formulas (e.g., ∫ 1/x dx = ln|x|, ∫ 1/(1+x^2) dx = tan⁻¹(x), ∫ e^x dx = e^x, ∫ sin(x) dx = -cos(x), etc.). Sometimes, integration by substitution might be required, as seen in Example 2.
4. Implicit vs. Explicit Solutions: Understand that sometimes it might not be possible (or necessary) to express y explicitly as a function of x. An implicit relation between x and y is perfectly acceptable as a general solution if it cannot be simplified further to y = f(x).

Practice Questions with Solutions

• Q: Solve the differential equation dy/dx = (4x+2)/y. A: Step 1: Separate the variables. y dy = (4x+2) dx Step 2: Integrate both sides. ∫ y dy = ∫ (4x+2) dx (y^2)/2 = 4(x^2)/2 + 2x + C (y^2)/2 = 2x^2 + 2x + C Step 3: Simplify for the general solution. y^2 = 4x^2 + 4x + 2C Let 2C = K. y^2 = 4x^2 + 4x + K Final answer: y^2 = 4x^2 + 4x + K
• Q: Find the particular solution of dy/dx = -4xy^2 given that y=1 when x=0. A: Step 1: Separate the variables. dy / y^2 = -4x dx Step 2: Integrate both sides. ∫ y^-2 dy = ∫ -4x dx -1/y = -4(x^2)/2 + C -1/y = -2x^2 + C Step 3: Use the initial condition y=1 when x=0 to find C. -1/1 = -2(0)^2 + C -1 = C Step 4: Substitute C back into the general solution for the particular solution. -1/y = -2x^2 - 1 1/y = 2x^2 + 1 y = 1 / (2x^2 + 1) Final answer: y = 1 / (2x^2 + 1)
• Q: Solve the differential equation (x^2 - yx^2)dy + (y^2 - xy^2)dx = 0. A: Step 1: Factorize common terms. x^2(1-y)dy + y^2(1-x)dx = 0 Step 2: Rearrange and separate variables. x^2(1-y)dy = -y^2(1-x)dx (1-y)/y^2 dy = -(1-x)/x^2 dx (1/y^2 - 1/y) dy = -(1/x^2 - 1/x) dx (y^-2 - y^-1) dy = (-x^-2 + x^-1) dx Step 3: Integrate both sides. ∫ (y^-2 - y^-1) dy = ∫ (-x^-2 + x^-1) dx -y^-1 - ln|y| = -(-x^-1) + ln|x| + C -1/y - ln|y| = 1/x + ln|x| + C Step 4: Rearrange for the general solution. 1/x + 1/y + ln|x| + ln|y| + C = 0 1/x + 1/y + ln|xy| + C = 0 Final answer: 1/x + 1/y + ln|xy| + C = 0
• Q: Solve the differential equation e^x tan(y) dx + (1 - e^x) sec^2(y) dy = 0. A: Step 1: Rearrange and separate variables. (1 - e^x) sec^2(y) dy = -e^x tan(y) dx sec^2(y) / tan(y) dy = -e^x / (1 - e^x) dx Step 2: Integrate both sides. ∫ (sec^2(y) / tan(y)) dy = ∫ (-e^x / (1 - e^x)) dx For LHS, let u = tan(y), then du = sec^2(y)dy. ∫ du/u = ln|u| = ln|tan(y)|. For RHS, let v = 1 - e^x, then dv = -e^x dx. ∫ dv/v = ln|v| = ln|1 - e^x|. So, ln|tan(y)| = ln|1 - e^x| + C Step 3: Express C in terms of logarithm for simplification. Let C = ln|K| where K is a constant. ln|tan(y)| = ln|1 - e^x| + ln|K| ln|tan(y)| = ln|K(1 - e^x)| Step 4: Remove logarithms. tan(y) = K(1 - e^x) Final answer: tan(y) = K(1 - e^x)

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