If y=a cos 3 x+b e^{-x},then y^{prime prime}( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $y = a \cos 3x + b e^{-x}$.First derivative: $y' = -3a \sin 3x - b e^{-x}$.Second derivative: $y'' = -9a \cos 3x + b e^{-x}$.Now,calculate $

Vedclass · Accepted Answer

$10 y^{\prime} \cos 3 x+3 y(\sin 3 x+3 \cos 3 x)$

Vedclass · Answer

(B) Given $y = a \cos 3x + b e^{-x}$.First derivative: $y' = -3a \sin 3x - b e^{-x}$.Second derivative: $y'' = -9a \cos 3x + b e^{-x}$.Now,calculate $y''(3 \sin 3x - \cos 3x)$:$y''(3 \sin 3x - \cos 3x) = (-9a \cos 3x + b e^{-x})(3 \sin 3x - \cos 3x)$$= -27a \cos 3x \sin 3x + 9a \cos^2 3x + 3b e^{-x} \sin 3x - b e^{-x} \cos 3x$.We can rewrite this expression by manipulating the terms to match the form $10y' \cos 3x + 3y(\sin 3x + 3 \cos 3x)$:$= 10(-3a \sin 3x - b e^{-x}) \cos 3x + 3a \cos 3x(\cos 3x + 3 \sin 3x) + 3b e^{-x}(\sin 3x + 3 \cos 3x)$$= 10y' \cos 3x + 3(a \cos 3x + b e^{-x})(\sin 3x + 3 \cos 3x)$$= 10y' \cos 3x + 3y(\sin 3x + 3 \cos 3x)$.

If $y=a \cos 3 x+b e^{-x}$,then $y^{\prime \prime}(3 \sin 3 x-\cos 3 x)=$

Similar Questions

If ${x^p}{y^q} = {(x + y)^{p + q}}$,then $\frac{{{d^2}y}}{{d{x^2}}} = $

If $y = at^2 + 2bt + c$ and $t = ax^2 + 2bx + c$,then $\frac{d^3y}{dx^3}$ equals

If $y = \frac{1}{x} + \cos 2x$,then $\frac{d^2y}{dx^2}$ is equal to

If $y = A \cos(nx) + B \sin(nx)$,then $\frac{d^2y}{dx^2} = $

If $y = \sin(m \sin^{-1} x)$,then $(1-x^2) y_2 - x y_1$ is equal to (Here,$y_n$ denotes $\frac{d^n y}{dx^n}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module