Class 12 Continuity And Differentiability · Long Answer3 Marksparametric_derivatives
Class 12 Maths · Continuity And Differentiability · Long Answer #100
If x=sin t and y=sin(pt), prove (1−x²)y″−xy′+p²y=0 wherever cos t≠0.
✅ Step-by-Step Solution
Final Answer: Identity proved.
- The coordinate derivatives are dx/dt=cos t and dy/dt=p cos(pt), so y′=p cos(pt)/cos t.
- Apply d/dx=(1/cos t)d/dt: y″=[−p²sin(pt)cos t+p cos(pt)sin t]/cos³t.
- Substitute 1−x²=cos²t, x=sin t and y=sin(pt). The p cos(pt)sin t/cos t terms cancel, leaving −p²sin(pt)+p²sin(pt)=0.
- Thus (1−x²)y″−xy′+p²y=0, as required.
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] The coordinate derivatives are dx/dt=cos t and dy/dt=p cos(pt), so y′=p cos(pt)/cos t.
- [1 mark] Apply d/dx=(1/cos t)d/dt: y″=[−p²sin(pt)cos t+p cos(pt)sin t]/cos³t.
- [1 mark] Substitute 1−x²=cos²t, x=sin t and y=sin(pt). The p cos(pt)sin t/cos t terms cancel, leaving −p²sin(pt)+p²sin(pt)=0.
- [0.5 marks] Thus (1−x²)y″−xy′+p²y=0, as required.
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