The differential equation obtained by elimina | vedclass

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

Question

(A) Given the equation: $y = A \cos \omega t + B \sin \omega t$ Differentiating with respect to $t$: $y' = -A \omega \sin \omega t + B \omega \cos \om

Vedclass · Accepted Answer

$y'' = - \omega^2 y$

Vedclass · Answer

(A) Given the equation: $y = A \cos \omega t + B \sin \omega t$ Differentiating with respect to $t$: $y' = -A \omega \sin \omega t + B \omega \cos \omega t$ Differentiating again with respect to $t$: $y'' = -A \omega^2 \cos \omega t - B \omega^2 \sin \omega t$ Factoring out $-\omega^2$: $y'' = -\omega^2 (A \cos \omega t + B \sin \omega t)$ Since $y = A \cos \omega t + B \sin \omega t$,we substitute $y$ into the equation: $y'' = -\omega^2 y$ Thus,the differential equation is $y'' + \omega^2 y = 0$.

The differential equation obtained by eliminating $A$ and $B$ from the equation $y = A \cos \omega t + B \sin \omega t$ is

Similar Questions

The differential equation whose solution is $y = c_{1} \cos(ax) + c_{2} \sin(ax)$ (where $c_{1}$ and $c_{2}$ are arbitrary constants) is

If the differential equation obtained by eliminating $A$ and $B$ from $y = (\sin^{-1} x)^2 + A \cos^{-1} x + B$ is $(a - x^2) y'' - x y' = b$,then $\frac{b + a}{b - a} =$

The differential equation corresponding to the family of ellipses $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$, where '$a$' is an arbitrary constant, is:

The differential equation of all circles which pass through the origin and whose centre lie on the $Y$-axis is

The differential equation whose solution is $y = A\sin x + B\cos x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module