Exercise 9.2, Question 1
\(y = e^x + 1\) : \(y'' - y' = 0\)
Show solution
- \(y' = e^x\), \(y'' = e^x\), so \(y'' - y' = 0\).
Answer: Verified
NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.2
Exercise 9.2: Verifying solutions. Differentiate the given function (implicitly if needed) as many times as the order of the equation, substitute into the left side, and simplify to the right side. For an implicit solution, find \(y'\) by implicit differentiation and use the original relation to simplify.
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Exercise 9.2 questions and solutions
Exercise 9.2, Question 2
\(y = x^2 + 2x + C\) : \(y' - 2x - 2 = 0\)
Show solution
- \(y' = 2x + 2\).
Answer: Verified
Exercise 9.2, Question 3
\(y = \cos x + C\) : \(y' + \sin x = 0\)
Show solution
- \(y' = -\sin x\).
Answer: Verified
Exercise 9.2, Question 4
\(y = \sqrt{1 + x^2}\) : \(y' = \dfrac{xy}{1 + x^2}\)
Show solution
- \[\begin{aligned}y' &= \dfrac{x}{\sqrt{1 + x^2}} \\ &= \dfrac{x\sqrt{1 + x^2}}{1 + x^2} \\ &= \dfrac{xy}{1 + x^2}\end{aligned}\]
Answer: Verified
Exercise 9.2, Question 5
\(y = Ax\) : \(xy' = y\) (\(x \ne 0\))
Show solution
- \(y' = A\), so \(xy' = Ax = y\).
Answer: Verified
Exercise 9.2, Question 6
\(y = x\sin x\) : \(xy' = y + x\sqrt{x^2 - y^2}\) (\(x \ne 0\), \(x > y\) or \(x < -y\))
Show solution
- \(y' = \sin x + x\cos x\), so \[\begin{aligned}xy' &= x\sin x + x^2\cos x \\ &= y + x^2\cos x\end{aligned}\]
- \[\begin{aligned}x^2 - y^2 &= x^2(1 - \sin^2 x) \\ &= x^2\cos^2 x\end{aligned}\], so \(x\sqrt{x^2 - y^2} = x\,|x\cos x|\), which is \(x^2\cos x\) wherever \(x\cos x > 0\) (the book's stated conditions do not quite guarantee this, but it is the case the verification intends).
Answer: Verified
Exercise 9.2, Question 7
\(xy = \log y + C\) : \(y' = \dfrac{y^2}{1 - xy}\) (\(xy \ne 1\))
Show solution
- Differentiate: \(y + xy' = \dfrac{y'}{y}\), so \(y'\left(\dfrac1y - x\right) = y\).
- \[\begin{aligned}y' &= \dfrac{y}{\frac{1 - xy}{y}} \\ &= \dfrac{y^2}{1 - xy}\end{aligned}\]
Answer: Verified
Exercise 9.2, Question 8
\(y - \cos y = x\) : \((y\sin y + \cos y + x)y' = y\)
Show solution
- \(y' + \sin y\,y' = 1\), so \(y' = \dfrac{1}{1 + \sin y}\).
- With \(x = y - \cos y\): \[\begin{aligned}y\sin y + \cos y + x &= y\sin y + y \\ &= y(1 + \sin y)\end{aligned}\]; times \(y'\) gives y.
Answer: Verified
Exercise 9.2, Question 9
\(x + y = \tan^{-1}y\) : \(y^2y' + y^2 + 1 = 0\)
Show solution
- \(1 + y' = \dfrac{y'}{1 + y^2}\), so \[y'\left(\dfrac{1}{1 + y^2} - 1\right) = 1\], i.e. \(y' = -\dfrac{1 + y^2}{y^2}\).
- \(y^2y' = -(1 + y^2)\).
Answer: Verified
Exercise 9.2, Question 10
\(y = \sqrt{a^2 - x^2}\), \(x \in (-a, a)\) : \(x + y\dfrac{dy}{dx} = 0\) (\(y \ne 0\))
Show solution
- \[\begin{aligned}y' &= -\dfrac{x}{\sqrt{a^2 - x^2}} \\ &= -\dfrac{x}{y}\end{aligned}\]
Answer: Verified
Exercise 9.2, Question 11
The number of arbitrary constants in the general solution of a fourth order equation is: (A) 0 (B) 2 (C) 3 (D) 4
Show solution
- The general solution of an order-n equation has n arbitrary constants.
Answer: (D)
Exercise 9.2, Question 12
The number of arbitrary constants in a particular solution of a third order equation is: (A) 3 (B) 2 (C) 1 (D) 0
Show solution
- A particular solution has all constants fixed.
Answer: (D)
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