Class 12 Continuity And Differentiability · Long Answer3 Markssecond_derivatives

Class 12 Maths · Continuity And Differentiability · Long Answer #90

For y=\(P e^{a x} + Q e^{b x}\), prove \(1\)y″ + \(- a - b\)y′ + \(a b\)y = \(0\). Assume all open intervals where the displayed expressions are real and differentiable.

✅ Step-by-Step Solution

Final Answer: Identity proved.

Differentiate y=\(P e^{a x} + Q e^{b x}\) to obtain y′=\(P a e^{a x} + Q b e^{b x}\).
Differentiate again, using the product and chain rules: y″=\(P a^{2} e^{a x} + Q b^{2} e^{b x}\).
Substitute these derivatives into the left side \(1\)y″ + \(- a - b\)y′ + \(a b\)y = \(0\). Collecting like terms gives \(0\).
This equals \(0\), proving the identity on the stated domain.

📋 CBSE Stepwise Marking Scheme

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

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Class 12 Maths · Continuity And Differentiability · Long Answer #90

✅ Step-by-Step Solution

📋 CBSE Stepwise Marking Scheme

Master Chapter 5: Continuity And Differentiability

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