If y=A cos n x+B sin n x, then y_2+n^2 y is e | vedclass

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

Question

(A) Given the function: $y = A \cos n x + B \sin n x$. First, differentiate with respect to $x$: $y_1 = \frac{dy}{dx} = -A n \sin n x + B n \cos n x$.

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given the function: $y = A \cos n x + B \sin n x$. First, differentiate with respect to $x$: $y_1 = \frac{dy}{dx} = -A n \sin n x + B n \cos n x$. Now, differentiate again with respect to $x$ to find the second derivative $y_2$: $y_2 = \frac{d^2y}{dx^2} = -A n^2 \cos n x - B n^2 \sin n x$. Factor out $-n^2$: $y_2 = -n^2 (A \cos n x + B \sin n x)$. Since $y = A \cos n x + B \sin n x$, we can substitute $y$ into the equation: $y_2 = -n^2 y$. Rearranging the terms gives: $y_2 + n^2 y = 0$.

If $y=A \cos n x+B \sin n x$, then $y_2+n^2 y$ is equal to

Similar Questions

The average degree of freedom per molecule of a gas is $6$. The gas performs $25 \, J$ of work while expanding at constant pressure. The heat absorbed by the gas is ...... $J$.

$A$ hollow sphere of radius $R$ is filled completely with an ideal liquid of density $\rho$. The sphere is moving horizontally with an acceleration $2g$,where $g$ is the acceleration due to gravity. If the minimum pressure of the liquid is $P_0$,then find the pressure at the centre of the sphere.

Among the following complexes,the one which shows zero Crystal Field Stabilization Energy $(CFSE)$ is

Consider the circle $x^2+y^2-4x-2y+c=0$ whose centre is $A(2,1)$. If the point $P(10,7)$ is such that the line segment $PA$ meets the circle at $Q$ with $PQ=5$,then $c$ is equal to

$A$ micelle formed during the cleansing action by soap is

Vedclass Test Series

Exam Paper Generator

Online Exam Module