Differential Equations Ex 9.2 Class 12 NCERT Solutions

What Does it Mean to Verify a Solution?

A differential equation is an equation that connects a function with its derivatives. For example, dy/dx = 2x is a simple differential equation. A solution to this equation is a function y(x) that makes the equation true. In this case, y = x^2 is a solution because if we differentiate it, we get dy/dx = 2x, which perfectly matches the equation.

Exercise 9.2 is not about finding this solution from scratch. Instead, you are given both the potential solution (e.g., y = x^2 + 3) and the differential equation (e.g., dy/dx = 2x). Your task is to verify if the function is indeed a solution. This is done by a straightforward process:

1. Find the necessary derivatives of the given function (like y' and y'').
2. Substitute the function and its derivatives back into the differential equation.
3. Simplify the equation. If the Left Hand Side (LHS) equals the Right Hand Side (RHS), the verification is complete, and the function is a solution. If they are not equal, it is not a solution.

How to Verify a Solution: A Step-by-Step Guide

1. Step 1: Identify the Function and the Equation — Clearly write down the given function, y = f(x), and the differential equation you need to check against.
2. Step 2: Differentiate the Function — Look at the differential equation to see the highest order of derivative required (e.g., y', y''). Calculate these derivatives from the given function y = f(x). Remember to use the chain rule, product rule, and quotient rule correctly.
3. Step 3: Substitute into the Differential Equation — Take the expressions for y, y', y'', etc., that you found and plug them into the differential equation. Usually, you substitute them into the more complex side (often the LHS).
4. Step 4: Simplify and Compare — Perform algebraic simplification on the expression you created in the previous step. After simplifying, check if the result is identical to the other side of the equation. If LHS = RHS, the function is a verified solution. Otherwise, it is not.

Worked Examples from NCERT Ex 9.2

• Question: Verify that the function y = e^(-3x) is a solution of the differential equation d²y/dx² + dy/dx - 6y = 0. Step 1: Given Function and DE Function: y = e^(-3x) Differential Equation: y'' + y' - 6y = 0 Step 2: Differentiate the Function First derivative (y'): y' = d/dx(e^(-3x)) = e^(-3x) (-3) = -3e^(-3x) (using chain rule) Second derivative (y''): y'' = d/dx(-3e^(-3x)) = -3 d/dx(e^(-3x)) = -3 (e^(-3x) -3) = 9e^(-3x) Step 3: Substitute into the LHS of the DE LHS = y'' + y' - 6y Substitute the expressions for y, y', and y'': LHS = (9e^(-3x)) + (-3e^(-3x)) - 6(e^(-3x)) Step 4: Simplify and Compare LHS = 9e^(-3x) - 3e^(-3x) - 6e^(-3x) LHS = (9 - 3 - 6)e^(-3x) LHS = 0 * e^(-3x) LHS = 0 Since LHS = 0 and RHS = 0, we have LHS = RHS. Conclusion: Hence, y = e^(-3x) is a solution of the given differential equation.
• Question: Verify that the function y = x sin(x) is a solution of the differential equation xy' = y + x sqrt(x² - y²). Step 1: Given Function and DE Function: y = x sin(x) Differential Equation: xy' = y + x sqrt(x² - y²) Step 2: Differentiate the Function We need y'. We use the product rule (uv)' = u'v + uv' where u=x and v=sin(x). y' = d/dx(x sin(x)) = (1)(sin(x)) + (x)(cos(x)) = sin(x) + x cos(x) Step 3: Substitute into LHS and RHS separately Let's evaluate the LHS of the DE: LHS = xy' = x(sin(x) + x cos(x)) = x sin(x) + x² cos(x) Now, let's evaluate the RHS of the DE: RHS = y + x sqrt(x² - y²) Substitute y = x sin(x) into the RHS: RHS = x sin(x) + x sqrt(x² - (x sin(x))²) RHS = x sin(x) + x sqrt(x² - x² sin²(x)) Step 4: Simplify and Compare RHS = x sin(x) + x sqrt(x²(1 - sin²(x))) Using the identity cos²(x) = 1 - sin²(x): RHS = x sin(x) + x sqrt(x² cos²(x)) RHS = x sin(x) + x (x cos(x)) (Assuming x > 0, cos(x) > 0 for sqrt) RHS = x sin(x) + x² cos(x) Comparing both sides, we find that LHS = x sin(x) + x² cos(x) and RHS = x sin(x) + x² cos(x). Conclusion: Since LHS = RHS, the function y = x sin(x) is a solution to the given differential equation.

Exam Traps and Common Mistakes

Students often lose marks in these seemingly simple verification problems due to silly mistakes. Here's what to watch out for:

• Differentiation Errors: The most common source of error. Double-check your application of the chain rule, product rule, and quotient rule. Forgetting to multiply by the derivative of the inner function (chain rule) is a frequent mistake (e.g., d/dx(cos(4x)) is -4sin(4x), not -sin(4x)).
• Algebraic Simplification: After substituting, the algebra can get messy. Be patient and systematic. A small sign error can make the entire verification fail.
• Substitution Mix-up: Be careful to substitute the correct expressions for y, y', and y''. It's helpful to write them out clearly before you start substituting.
• Implicit vs. Explicit Differentiation: For functions where y is not easily isolated, you might need to use implicit differentiation. Ensure you're comfortable with this technique.

Practice Questions with Solutions

• Q: Verify that the function y = x² + 2x + C is a solution of the differential equation y' - 2x - 2 = 0. A: Step 1: Find the derivative of the given function. y = x² + 2x + C y' = d/dx(x² + 2x + C) = 2x + 2 Step 2: Substitute y' into the Left Hand Side (LHS) of the differential equation. LHS = y' - 2x - 2 LHS = (2x + 2) - 2x - 2 Step 3: Simplify the expression. LHS = 2x - 2x + 2 - 2 = 0 Step 4: Compare LHS with RHS. LHS = 0 and RHS = 0. So, LHS = RHS. Final answer: Hence, y = x² + 2x + C is a solution to the given differential equation.
• Q: Verify that y = A cos(x) - B sin(x) is a solution of the differential equation y'' + y = 0. A: Step 1: Find the first and second derivatives of the function. y = A cos(x) - B sin(x) y' = -A sin(x) - B cos(x) y'' = -A cos(x) - B(-sin(x)) = -A cos(x) + B sin(x) Step 2: Substitute the expressions for y'' and y into the LHS of the DE. LHS = y'' + y LHS = (-A cos(x) + B sin(x)) + (A cos(x) - B sin(x)) Step 3: Simplify the LHS. LHS = -A cos(x) + A cos(x) + B sin(x) - B sin(x) = 0 Step 4: Compare LHS with RHS. LHS = 0 and RHS = 0. So, LHS = RHS. Final answer: Hence, y = A cos(x) - B sin(x) is a solution to the given differential equation.
• Q: Verify that y = log(x) is a solution to the differential equation xy'' + y' = 0. A: Step 1: Find the first and second derivatives of the function y = log(x). y' = d/dx(log(x)) = 1/x y'' = d/dx(1/x) = d/dx(x⁻¹) = -1 * x⁻² = -1/x² Step 2: Substitute y' and y'' into the LHS of the differential equation. LHS = xy'' + y' LHS = x(-1/x²) + (1/x) Step 3: Simplify the expression. LHS = -1/x + 1/x = 0 Step 4: Compare LHS with RHS. LHS = 0 and RHS = 0. So, LHS = RHS. Final answer: Hence, y = log(x) is a solution to the given differential equation.
• Q: Is the function y = x³ a solution to the differential equation xy' - 3y = 0? A: Step 1: Find the derivative of the given function. y = x³ y' = 3x² Step 2: Substitute y and y' into the LHS of the differential equation. LHS = xy' - 3y LHS = x(3x²) - 3(x³) Step 3: Simplify the expression. LHS = 3x³ - 3x³ = 0 Step 4: Compare LHS with RHS. LHS = 0 and RHS = 0. Thus, LHS = RHS. Final answer: Yes, y = x³ is a solution to the given differential equation.

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