Second Order Derivatives PYQs
Practice Class 12 CBSE Board Previous Year Questions (2014-2026)
Q1
2024
00:00
If $y = 3e^{2x} + 2e^{3x}$, prove $y'' - 5y' + 6y = 0$. 3 Marks
$y' = 6e^{2x} + 6e^{3x}$.
$y'' = 12e^{2x} + 18e^{3x}$.
Substitute in LHS: $(12e^{2x} + 18e^{3x}) - 5(6e^{2x} + 6e^{3x}) + 6(3e^{2x} + 2e^{3x})$.
$= (12 - 30 + 18)e^{2x} + (18 - 30 + 12)e^{3x} = 0 + 0 = 0$.
Proved
Q2
2023
00:00
If $y = A \sin x + B \cos x$, prove $y'' + y = 0$. 2 Marks
$y' = A \cos x - B \sin x$.
$y'' = -A \sin x - B \cos x$.
$y'' = -(A \sin x + B \cos x) = -y \Rightarrow y'' + y = 0$.
Proved
Q3
2022 (Term 2)
00:00
If $y = \sin^{-1} x$, show that $(1-x^2)y'' - xy' = 0$. 3 Marks
$y' = \frac{1}{\sqrt{1-x^2}} \Rightarrow \sqrt{1-x^2} y' = 1$.
Differentiate again: $\sqrt{1-x^2} y'' + y' \frac{-x}{\sqrt{1-x^2}} = 0$.
Multiply by $\sqrt{1-x^2}$: $(1-x^2)y'' - xy' = 0$.
Proved
Q4
2020
00:00
If $y = (\sin^{-1} x)^2$, prove $(1-x^2)y'' - xy' - 2 = 0$. 4 Marks
$y' = 2(\sin^{-1} x) \frac{1}{\sqrt{1-x^2}} \Rightarrow (1-x^2) (y')^2 = 4(\sin^{-1} x)^2 = 4y$.
Differentiate w.r.t $x$: $(1-x^2) 2y' y'' + (y')^2(-2x) = 4y'$.
Divide by $2y'$: $(1-x^2) y'' - x y' = 2 \Rightarrow (1-x^2) y'' - x y' - 2 = 0$.
Proved
Q5
2019
00:00
If $y = e^{a \cos^{-1} x}, -1 \le x \le 1$, show that $(1 - x^2) y'' - x y' - a^2 y = 0$. 5 Marks
$y' = e^{a \cos^{-1} x} \frac{-a}{\sqrt{1-x^2}} = \frac{-ay}{\sqrt{1-x^2}}$.
$(1-x^2)(y')^2 = a^2 y^2$.
Differentiate again: $(1-x^2) 2y' y'' + (y')^2(-2x) = a^2 2y y'$.
Divide by $2y'$: $(1-x^2) y'' - x y' = a^2 y$.
Proved
Q6
2015
00:00
If $y = A e^{mx} + B e^{nx}$, prove $y'' - (m+n)y' + mn \cdot y = 0$. 5 Marks
$y' = Am e^{mx} + Bn e^{nx}$.
$y'' = Am^2 e^{mx} + Bn^2 e^{nx}$.
LHS $= (Am^2 e^{mx} + Bn^2 e^{nx}) - (m+n)(Am e^{mx} + Bn e^{nx}) + mn(A e^{mx} + B e^{nx})$.
Coeff of $A e^{mx}: m^2 - (m+n)m + mn = m^2 - m^2 - mn + mn = 0$.
Coeff of $B e^{nx}: n^2 - (m+n)n + mn = n^2 - mn - n^2 + mn = 0$.
So LHS $= 0$.
Proved
Q7
2017
00:00
If $y = e^x(\sin x + \cos x)$, prove $y'' - 2y' + 2y = 0$. 3 Marks
$y' = e^x(\cos x - \sin x) + e^x(\sin x + \cos x) = 2e^x \cos x$.
$y'' = 2e^x(\cos x - \sin x)$.
LHS $= 2e^x \cos x - 2e^x \sin x - 2(2e^x \cos x) + 2(e^x \sin x + e^x \cos x)$.
$= (2 - 4 + 2)e^x \cos x + (-2 + 2)e^x \sin x = 0 + 0 = 0$.
Proved
Q8
2016
00:00
If $y = 3 \cos(\log x) + 4 \sin(\log x)$, show that $x^2 y'' + x y' + y = 0$. 5 Marks
$y' = -3\sin(\log x) \frac{1}{x} + 4\cos(\log x) \frac{1}{x} \Rightarrow x y' = -3\sin(\log x) + 4\cos(\log x)$.
Differentiate again w.r.t $x$: $x y'' + y'(1) = -3\cos(\log x) \frac{1}{x} - 4\sin(\log x) \frac{1}{x}$.
$x^2 y'' + x y' = -(3\cos(\log x) + 4\sin(\log x)) = -y$.
$x^2 y'' + x y' + y = 0$.
Proved