The differential equation, representing the curve y=e^x(a x+b x) where a and b are arbitrary constants, is:

Correct answer: C. (d^2 y)/(d x^2)-2 (d y)/(d x)+2 y=0 y=e^x(a x+b x) ...(i) aligned (d y)/(d x)= & e^x(-a x+b x) & +(a x+b x) e^x (d y)/(d x)= & e^x(-a x+b x)+y aligned Again differentiate w.r.t. to x aligned (d^2 y)/(d x^2)= & -e^x(a x+b x) & +(-a x+b x) e^x+(d y)/(d x) (d^2 y)/(d x^2)= & -y+(d y)/(d x)-y+(d y)/(d x)[ using (i) and (ii) ] (d^2 y)/(d x^2)- & 2 (d y)/(d x)+2 y=0 aligned is the required differential equation.

NDA 2024

Exam: NDA
Year: 2024
Subject: Mathematics
Chapter: Differential Equations

NDA 2024 · Mathematics · Differential Equations

The differential equation, representing the curve y=e^x(a x+b x)...

The differential equation, representing the curve

y=e^x(a \cos x+b \sin x)

where

a

and

b

are arbitrary constants, is:

Options

A $\frac{d^2 y}{d x^2}+2 y=0$
B $\frac{d^2 y}{d x^2}+2 \frac{d y}{d x}+2 y=0$
C $\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+2 y=0$
D $\frac{d^2 y}{d x^2}+y=0$

Correct answer

C. $\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+2 y=0$

Step-by-step solution

y=e^x(a \cos x+b \sin x)\quad ...(i)

\begin{aligned} \frac{d y}{d x}= & e^x(-a \sin x+b \cos x) \\ & +(a \cos x+b \sin x) e^x \\ \frac{d y}{d x}= & e^x(-a \sin x+b \cos x)+y \end{aligned}

Again differentiate w.r.t. to $x

\begin{aligned} \frac{d^2 y}{d x^2}= & -e^x(a \cos x+b \sin x) \\ & +(-a \sin x+b \cos x) e^x+\frac{d y}{d x} \\ \frac{d^2 y}{d x^2}= & -y+\frac{d y}{d x}-y+\frac{d y}{d x}[\text{ using (i) and (ii) }] \\ \frac{d^2 y}{d x^2}- & 2 \frac{d y}{d x}+2 y=0 \end{aligned}$$$ is the required differential equation.

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