More practice questions

5. Q5. Verify that y = x² + 2x + C is a solution of y′ − 2x − 2 = 0 (NCERT Exercise 9.2, Q2, page 306).

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   y′ = 2x + 2, so y′ − 2x − 2 = 0 for every value of C.

6. Q6. Verify that y = √(1 + x²) is a solution of y′ = xy/(1 + x²) (NCERT Exercise 9.2, Q4, page 306).

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   y′ = x/√(1 + x²). Since y/(1 + x²) = 1/√(1 + x²), this is xy/(1 + x²).

7. Q7. Verify that y − cos y = x is a solution of (y sin y + cos y + x) y′ = y (NCERT Exercise 9.2, Q8, page 306).

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   Work on a branch where 1 + sin y ≠ 0; otherwise the inverse relation does not give a finite derivative. Differentiate: y′ + sin y · y′ = 1, so y′ = 1/(1 + sin y). In the bracket put x = y − cos y: it becomes y sin y + y, that is, y(1 + sin y). Times y′ this gives y.

8. Q8. Verify that x + y = tan⁻¹ y is a solution of y²y′ + y² + 1 = 0 (NCERT Exercise 9.2, Q9, page 306).

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   Work where y ≠ 0: at y = 0 the proposed differential equation would read 1 = 0, so the implicit curve is not a differentiable solution there. Differentiate: 1 + y′ = y′/(1 + y²). Multiply by 1 + y²: 1 + y² + (1 + y²)y′ = y′, so y²y′ + y² + 1 = 0.

9. Q9. Verify that y = √(a² − x²), for a > 0 and x in (−a, a), is a solution of x + y dy/dx = 0 (NCERT Exercise 9.2, Q10, page 306).

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   dy/dx = −x/√(a² − x²), which is −x/y. So y dy/dx = −x, and x + y dy/dx = 0.

10. Q10. Beyond the NCERT text: from the general solution y = a cos x + b sin x of y″ + y = 0 (Example 2), find the particular solution with y = 1 and y′ = 2 when x = 0.

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    At x = 0, y = a, so a = 1; and y′ = −a sin x + b cos x gives y′ = b, so b = 2. The particular solution is y = cos x + 2 sin x. Two conditions fix the two constants.