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Class 12 Continuity And Differentiability · Long Answer3 Markssecond_derivatives

Class 12 Maths · Continuity And Differentiability · Long Answer #88

For y=\(e^{a \operatorname{acos}{\left(x \right)}}\), prove \(1 - x^{2}\)y″ + \(- x\)y′ + \(- a^{2}\)y = \(0\). Assume −1<x<1 and a is a real constant.

✅ Step-by-Step Solution

Final Answer: Identity proved.
  1. Differentiate y=\(e^{a \operatorname{acos}{\left(x \right)}}\) to obtain y′=\(- \frac{a e^{a \operatorname{acos}{\left(x \right)}}}{\sqrt{1 - x^{2}}}\).
  2. Differentiate again, using the product and chain rules: y″=\(\frac{a \left(a \left(1 - x^{2}\right)^{\frac{3}{2}} + x \left(x^{2} - 1\right)\right) e^{a \operatorname{acos}{\left(x \right)}}}{\left(1 - x^{2}\right)^{\frac{5}{2}}}\).
  3. Substitute these derivatives into the left side \(1 - x^{2}\)y″ + \(- x\)y′ + \(- a^{2}\)y = \(0\). Collecting like terms gives \(0\).
  4. This equals \(0\), proving the identity on the stated domain.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Differentiate y=\(e^{a \operatorname{acos}{\left(x \right)}}\) to obtain y′=\(- \frac{a e^{a \operatorname{acos}{\left(x \right)}}}{\sqrt{1 - x^{2}}}\).
  • [1 mark] Differentiate again, using the product and chain rules: y″=\(\frac{a \left(a \left(1 - x^{2}\right)^{\frac{3}{2}} + x \left(x^{2} - 1\right)\right) e^{a \operatorname{acos}{\left(x \right)}}}{\left(1 - x^{2}\right)^{\frac{5}{2}}}\).
  • [1 mark] Substitute these derivatives into the left side \(1 - x^{2}\)y″ + \(- x\)y′ + \(- a^{2}\)y = \(0\). Collecting like terms gives \(0\).
  • [0.5 marks] This equals \(0\), proving the identity on the stated domain.

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