NCERT Class 12 Maths · Chapter 9 Differential Equations · Miscellaneous Exercise Question 2

✅ Step-by-Step Solution

1. Part 1: y′=\(\frac{a e^{x} - b e^{- x} + 2 x}{x} - \frac{a e^{x} + b e^{- x} + x^{2}}{x^{2}}\); y″=\(\frac{a e^{x} + b e^{- x} + 2 - \frac{2 \left(a e^{x} - b e^{- x} + 2 x\right)}{x} + \frac{2 \left(a e^{x} + b e^{- x} + x^{2}\right)}{x^{2}}}{x}\).
2. Substituting these into xy″+2y′−xy+x²−2 gives 0 after cancellation.
3. Part 2: y′=\(\left(- a \sin{\left(x \right)} + b \cos{\left(x \right)}\right) e^{x} + \left(a \cos{\left(x \right)} + b \sin{\left(x \right)}\right) e^{x}\); y″=\(- 2 \left(a \sin{\left(x \right)} - b \cos{\left(x \right)}\right) e^{x}\).
4. Substituting these into y″−2y′+2y gives 0 after cancellation.
5. Part 3: y′=\(3 x \cos{\left(3 x \right)} + \sin{\left(3 x \right)}\); y″=\(3 \left(- 3 x \sin{\left(3 x \right)} + 2 \cos{\left(3 x \right)}\right)\).
6. Substituting these into y″+9y−6cos 3x gives 0 after cancellation.
7. Part (iv): x²=2y²log y requires y>0. Differentiate: 2x=(4y log y+2y)y′.
8. Hence y′=x/[y(2log y+1)]. On the curve, x²+y²=y²(2log y+1), so (x²+y²)y′=xy.