If $y=e^{-x}(A\cos x+B\sin x)$, then $y$ is a solution of: (A) $\frac{d^2y}{dx^2}+2\frac{dy}{dx}=0$ (B) $\frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y=0$ (C) $\frac{d^2y}{dx^2}+2\frac{dy}{dx}+2y=0$ (D) $\frac{d^2y}{dx^2}+2y=0$

If y=e-x(Acos x+Bsin x), then y is a solution of: (A) (d²y)/(dx²)+2(dy)/(dx)=0 (B) (d²y)/(dx²)-2(dy)/(dx)+2y=0

NCERT Exemplar · Q59

y=e^{-x}(A\cos x+B\sin x)

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The given function $y = e^{-x}(A\cos x + B\sin x)$ is a linear combination of $e^{-x}\cos x$ and $e^{-x}\sin x$ , which are solutions of the second-order linear ODE with characteristic roots $-1 \pm i$ . The corresponding differential equation is $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 2y = 0$ , which is option (C).

The core idea here is verification of a solution — we are given a candidate function and need to check which differential equation it satisfies. Instead of solving each ODE from scratch, we can compute the derivatives of $y$ and substitute them into each option. But there's a more elegant way: recognise the form.

The function $y = e^{-x}(A\cos x + B\sin x)$ is the general solution of a second-order linear homogeneous ODE with constant coefficients. The characteristic equation for such an ODE is $r^2 + pr + q = 0$ , and the solution form $e^{\alpha x}(C_1\cos\beta x + C_2\sin\beta x)$ corresponds to complex conjugate roots $\alpha \pm i\beta$ .

Here, $\alpha = -1$ and $\beta = 1$ . So the characteristic roots are $-1 \pm i$ . The characteristic equation is therefore:

(r - (-1 + i))(r - (-1 - i)) = 0

which simplifies to:

(r + 1 - i)(r + 1 + i) = (r+1)^2 + 1 = r^2 + 2r + 2 = 0

Thus the ODE is $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 2y = 0$ .

Let's verify this by direct differentiation as well.

First derivative:

$y = e^{-x}(A\cos x + B\sin x)$

Using the product rule:

$\frac{dy}{dx} = -e^{-x}(A\cos x + B\sin x) + e^{-x}(-A\sin x + B\cos x)$

Factor $e^{-x}$ :

$\frac{dy}{dx} = e^{-x}\left[ -A\cos x - B\sin x - A\sin x + B\cos x \right]$

Group $\cos x$ and $\sin x$ terms:

$\frac{dy}{dx} = e^{-x}\left[ (-A + B)\cos x + (-B - A)\sin x \right]$
Second derivative:

Differentiate $\frac{dy}{dx}$ again. Let $P = -A + B$ and $Q = -A - B$ , so $\frac{dy}{dx} = e^{-x}(P\cos x + Q\sin x)$ .

Then:

$\frac{d^2y}{dx^2} = -e^{-x}(P\cos x + Q\sin x) + e^{-x}(-P\sin x + Q\cos x)$

$= e^{-x}\left[ (-P + Q)\cos x + (-Q - P)\sin x \right]$

Substitute back $P$ and $Q$ :

$-P + Q = -(-A+B) + (-A-B) = A - B - A - B = -2B$

$-Q - P = -(-A-B) - (-A+B) = A + B + A - B = 2A$

So $\frac{d^2y}{dx^2} = e^{-x}(-2B\cos x + 2A\sin x)$
Now check option (C): $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 2y = 0$

Compute $2\frac{dy}{dx} = 2e^{-x}\left[ (-A + B)\cos x + (-A - B)\sin x \right]$

And $2y = 2e^{-x}(A\cos x + B\sin x)$

Add them: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-A1 markMCQ
$\boldsymbol{x}+\boldsymbol{y}=\boldsymbol{\operatorname { t a n }}^{-\mathbf{1}} \boldsymbol{y}$
›Reveal solutionSolution

Differentiating $x+y=\tan^{-1}y$ gives $y^2y'+y^2+1=0$ — option (C).

Differentiate the relation $x+y=\tan^{-1}y$ with respect to $x$ :
$1+y'=\frac{1}{1+y^2}\,y'$
Multiply both sides by $(1+y^2)$ :
$(1+y^2)+(1+y^2)y'=y'$ $(1+y^2)+y'+y^2y'-y'=0$
$1+y^2+y^2y'=0$ …
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KCET 2020Set A-11 markMCQ
$y = 2x^{n+1} + \frac{3}{x^n}$
›Reveal solutionSolution

Differentiate the power function twice, multiply by $x^2$ , and notice the result is $n(n+1)$ times the original $y$ .

Step 1 — Write $y$ with negative exponents (so the power rule applies to both terms).
$y = 2x^{n+1} + \frac{3}{x^{n}} = 2x^{n+1} + 3x^{-n}.$
Step 2 — First derivative (power rule $\frac{d}{dx}x^{m} = m x^{m-1}$ ):
$\frac{dy}{dx} = 2(n+1)x^{n} + 3(-n)x^{-n-1} = 2(n+1)x^{n} - 3n\,x^{-n-1}.$
Step 3 — Second derivative:
$\frac{d^{2}y}{dx^{2}} = 2(n+1)\,n\,x^{n-1} - 3n(-n-1)x^{-n-2} = 2n(n+1)x^{n-1} + 3n(n+1)x^{-n-2}.$
Step 4 — Multiply by $x^{2}$ and factor.

$x^{2}\frac{d^{2}y}{dx^{2}} = 2n(n+1)x^{n+1} + 3n(n+1)x^{-n} = n(n+1)\Big[\,2x^{n+1} + 3x^{-n}\,\Big] = n(n+1)\,y.$ …
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KCET 2021Set A-11 markMCQ
$\left(\frac{1+i}{1-i}\right)^x = 1$
›Reveal solutionSolution

Rationalise $\frac{1+i}{1-i}$ to get $i$ , then use the period-4 cycle of powers of $i$ : $i^x=1$ exactly when $x$ is a multiple of 4.

Step 1 — Simplify the base by rationalising the denominator.

The standard move for a complex fraction is to multiply top and bottom by the conjugate of the denominator. The conjugate of $1-i$ is $1+i$ :
$\frac{1+i}{1-i} = \frac{1+i}{1-i}\times\frac{1+i}{1+i} = \frac{(1+i)^2}{(1-i)(1+i)}.$
Step 2 — Expand numerator and denominator.

Numerator (using $i^2=-1$ ):
$(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i.$
Denominator (difference of squares — this is why we use the conjugate: it clears $i$ from the bottom):
$(1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 2.$
Therefore
$\frac{1+i}{1-i} = \frac{2i}{2} = i.$
Step 3 — Rewrite the equation.
$\left(\frac{1+i}{1-i}\right)^{x} = 1 \quad\Longrightarrow\quad i^{x} = 1.$
Step 4 — The concept: powers of $i$ are periodic with period 4.
$i^1 = i,\qquad i^2 = -1,\qquad i^3 = -i,\qquad i^4 = 1,$
and then the cycle repeats: $i^5=i$ , $i^6=-1$ , and so on. In general $i^{x}$ depends only on $x \bmod 4$ , and

$i^{x} = 1 \iff x \equiv 0 \pmod 4 \iff x = 4n.$ …
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KCET 2022Set C-41 markMCQ
$3x + i(4x - y) = 6 - i$
›Reveal solutionSolution

Equate real and imaginary parts on the two sides of the complex equation and solve the resulting pair of linear equations.

Step 1 — The concept: equality of complex numbers.

If $a+ib=c+id$ with $a,b,c,d\in\mathbb{R}$ , then $a=c$ and $b=d$ . This works because $\{1,i\}$ is a basis of $\mathbb{C}$ over $\mathbb{R}$ : a real number can never equal a non-zero purely imaginary number, so the two components cannot compensate for one another.

Step 2 — Write both sides in the standard $a+ib$ form.
$3x+i(4x-y)=6+i(-1).$
Here $x,y$ are real, so $3x$ is the real part on the left and $(4x-y)$ is the imaginary part.

Step 3 — Compare real parts. …
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KCET 2023Set A-21 markMCQ
$\dfrac{(1+i)^2 (1+3i)}{(2-6i)(2-2i)}$
›Reveal solutionSolution

Use $\left|\dfrac{z_1z_2}{z_3z_4}\right|=\dfrac{|z_1||z_2|}{|z_3||z_4|}$ — take moduli factor by factor instead of expanding the messy product.

Step 1 — Why this works.

The modulus is multiplicative ( $|z_1z_2|=|z_1||z_2|$ , $|z_1/z_2|=|z_1|/|z_2|$ ), so we never need to expand the complex arithmetic.

Step 2 — Numerator moduli.
$|1+i|=\sqrt{1^2+1^2}=\sqrt2\ \Rightarrow\ |(1+i)^2|=(\sqrt2)^2=2$ $|1+3i|=\sqrt{1^2+3^2}=\sqrt{10}$ $\text{Numerator}=2\sqrt{10}$
Step 3 — Denominator moduli.
$|2-6i|=\sqrt{4+36}=\sqrt{40}=2\sqrt{10}$ $|2-2i|=\sqrt{4+4}=\sqrt8=2\sqrt2$
$\text{Denominator}=2\sqrt{10}\cdot 2\sqrt2=4\sqrt{20}=8\sqrt5$ …
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KCET 2025Set A-11 markMCQ
$3x + 8 < 17$
›Reveal solutionSolution

Solve both inequalities (Statement I checks out), then note the common solution is the interval $[2,3)$ , not the two-element set $\{2,3\}$ — so Statement II is false.

Step 1 — Solve the first inequality.
$3x+8<17\ \Longrightarrow\ 3x<9\ \Longrightarrow\ x<3 .$
Step 2 — Solve the second inequality.
$2x+8\ge 12\ \Longrightarrow\ 2x\ge 4\ \Longrightarrow\ x\ge 2 .$
So Statement (I) — "the solution sets are $x<3$ and $x\ge2$ respectively" — is TRUE.

Step 3 — Find the common solution set.

Intersecting $x<3$ with $x\ge2$ :
$2\le x<3\qquad\text{i.e.}\qquad x\in[2,3).$
Step 4 — Why Statement (II) fails — two independent reasons.

Statement (II) claims the common set is $\{2,3\}$ . …
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KCET 2023Set A-21 markMCQ
$p\left(\frac{1}{q}\right), q\left(\frac{1}{r}\right), r\left(\frac{1}{p}\right), \left(\frac{1}{p}\right)\left(\frac{1}{q}\right)$
›Reveal solutionSolution

Add $2$ to each of the three terms; every term then factorises as $(p+q+r)\left(\frac1p+\frac1q+\frac1r\right)$ times $p$ , $q$ , $r$ respectively — so the given A.P. forces $p,q,r$ themselves to be in A.P.

Step 1 — The three terms

The terms are
$T_1 = p\left(\frac1q+\frac1r\right),\qquad T_2 = q\left(\frac1r+\frac1p\right),\qquad T_3 = r\left(\frac1p+\frac1q\right).$
Step 2 — Add a constant (A.P. is preserved)

If $T_1,T_2,T_3$ are in A.P., so are $T_1+2,\ T_2+2,\ T_3+2$ (adding the same constant to every term does not change the common difference).

Write $2 = 1 + 1$ and absorb one of the $1$ s as $\frac{p}{p}$ :
$T_1 + 2 = \frac{p}{q}+\frac{p}{r} + \frac{p}{p} + 1 = p\left(\frac1p+\frac1q+\frac1r\right)+1 .$
By the same symmetry,
$T_2+2 = q\left(\frac1p+\frac1q+\frac1r\right)+1,\qquad T_3+2 = r\left(\frac1p+\frac1q+\frac1r\right)+1 .$
Step 3 — Strip the constants

Subtract $1$ from each, then divide each by the common non-zero factor $k=\left(\frac1p+\frac1q+\frac1r\right)$ . Both operations preserve an A.P. We are left with
$p,\qquad q,\qquad r$
in A.P.

So the given terms are in A.P. iff $p,q,r$ are in A.P.

Step 4 — Numerical verification …
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KCET 2022Set C-41 markMCQ
$-1, 0, 1, k$
›Reveal solutionSolution

Apply $\sigma^{2}=\dfrac{\sum x_i^{2}}{n}-\bar{x}^{2}$ to the four numbers, set it equal to $(\sqrt5)^2=5$ , and solve the resulting quadratic in $k$ .

Step 1 — Set up the data

The observations are $-1,\;0,\;1,\;k$ , so $n=4$ .
$\sum x_i = -1+0+1+k = k \quad\Rightarrow\quad \bar{x}=\frac{k}{4}$ $\sum x_i^{2} = (-1)^{2}+0^{2}+1^{2}+k^{2} = 2+k^{2}$
Step 2 — Use the computational formula for variance

The formula $\sigma^{2}=\dfrac{\sum x_i^{2}}{n}-\bar{x}^{2}$ (mean of squares minus square of the mean) is the efficient route here, because $\bar x$ is not a whole number and the deviation form would be messy.
$\sigma^{2}=\frac{2+k^{2}}{4}-\left(\frac{k}{4}\right)^{2}=\frac{2+k^{2}}{4}-\frac{k^{2}}{16}$
Step 3 — Impose the given standard deviation

Given $\sigma=\sqrt5$ , so $\sigma^{2}=5$ :
$\frac{2+k^{2}}{4}-\frac{k^{2}}{16}=5$
Multiply throughout by $16$ (the LCM of the denominators):
$4\left(2+k^{2}\right)-k^{2}=80$
$8+4k^{2}-k^{2}=80$ …
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KCET 2018Set A-11 markMCQ
$Q. Everybody in a room shakes hands with everybody else. The total number of handshakes is 45. The total number of persons in the room is (A) 9 (B) 10 (C) 5 (D) 15$
›Reveal solutionSolution

Each handshake involves a pair of people, so the total is $^nC_2$ ; set $^nC_2 = 45$ and solve for $n$ .

Step 1 — Model the situation.

A handshake is completely determined by which two people shake — the order does not matter (A shaking B is the same handshake as B shaking A), and a person cannot shake their own hand. So the count of handshakes among $n$ people is a combination, not a permutation:
$\text{Number of handshakes} = {}^{n}C_{2} = \frac{n(n-1)}{2}$
Step 2 — Form and solve the equation.
$\frac{n(n-1)}{2} = 45 \;\Rightarrow\; n(n-1) = 90 \;\Rightarrow\; n^{2} - n - 90 = 0$
$\Rightarrow\; (n-10)(n+9) = 0 \;\Rightarrow\; n = 10 \ \text{or}\ n = -9$ …
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Q.If $y=e^{-x}(A\cos x+B\sin x)$ , then $y$ is a solution of:
(A) $\frac{d^2y}{dx^2}+2\frac{dy}{dx}=0$
(B) $\frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y=0$
(C) $\frac{d^2y}{dx^2}+2\frac{dy}{dx}+2y=0$
(D) $\frac{d^2y}{dx^2}+2y=0$

Concept understanding — Verification of Solution

Verifying a Solution of a Differential Equation

The steps

Example 1

Example 2 (a solution with constants)

Method: Testing Which DE a Given Function Satisfies

Steps

Common Mistakes

Step 1 — The three terms

Step 2 — Add a constant (A.P. is preserved)

Step 3 — Strip the constants

Step 4 — Numerical verification …

Step 1 — Set up the data

Step 2 — Use the computational formula for variance

Step 3 — Impose the given standard deviation

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