If y = a sin x + (5 + 2x) cos x,then y'' + y | vedclass

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

Question

(D) Given: $y = a \sin x + (5 + 2x) \cos x$ . . . . . . $(i)$Differentiating with respect to $x$,we get:$y' = a \cos x + 2 \cos x - (5 + 2x) \sin x$$y

Vedclass · Accepted Answer

$-4 \sin x$

Vedclass · Answer

(D) Given: $y = a \sin x + (5 + 2x) \cos x$ . . . . . . $(i)$Differentiating with respect to $x$,we get:$y' = a \cos x + 2 \cos x - (5 + 2x) \sin x$$y' = (a + 2) \cos x - (5 + 2x) \sin x$Again differentiating with respect to $x$,we get:$y'' = -(a + 2) \sin x - 2 \sin x - (5 + 2x) \cos x$$y'' = -(a + 4) \sin x - (5 + 2x) \cos x$ . . . . . . $(ii)$Adding $(i)$ and $(ii)$,we get:$y'' + y = [-(a + 4) \sin x - (5 + 2x) \cos x] + [a \sin x + (5 + 2x) \cos x]$$y'' + y = -a \sin x - 4 \sin x - (5 + 2x) \cos x + a \sin x + (5 + 2x) \cos x$$y'' + y = -4 \sin x$.

If $y = a \sin x + (5 + 2x) \cos x$,then $y'' + y =$

Similar Questions

If $\log _e y=3 \sin ^{-1} x$,then $(1-x^2) y^{\prime \prime}-x y^{\prime}$ at $x=\frac{1}{2}$ is equal to :

$f(x) = e^x \sin x$,then $f^{(6)}(x)$ is equal to :

If $y=e^{m \sin ^{-1} x}$, then $(1-x^{2}) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-k y=0$, where $k$ is equal to

If $y = (\sin^{-1} x)^2 + (\cos^{-1} x)^2$,then $(1 - x^2) y_2 - x y_1 =$

The equation $x^5 - 5x^3 + 5x^2 - 1 = 0$ has how many equal roots?

Vedclass Test Series

Exam Paper Generator

Online Exam Module