The differential equation whose solution is y | vedclass

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

Question

(A) Given the equation $y = A\sin x + B\cos x$. Differentiating with respect to $x$ once,we get: $\frac{dy}{dx} = A\cos x - B\sin x$. Differentiating

Vedclass · Accepted Answer

$\frac{d^2y}{dx^2} + y = 0$

Vedclass · Answer

(A) Given the equation $y = A\sin x + B\cos x$. Differentiating with respect to $x$ once,we get: $\frac{dy}{dx} = A\cos x - B\sin x$. Differentiating with respect to $x$ again,we get: $\frac{d^2y}{dx^2} = -A\sin x - B\cos x$. We can rewrite this as: $\frac{d^2y}{dx^2} = -(A\sin x + B\cos x)$. Since $y = A\sin x + B\cos x$,we substitute $y$ into the equation: $\frac{d^2y}{dx^2} = -y$. Rearranging the terms,we get the required differential equation: $\frac{d^2y}{dx^2} + y = 0$.

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

Similar Questions

The differential equation of all circles touching the $Y$-axis at the origin and having their center on the $X$-axis is:

Verify that the given function $x^{2}=2 y^{2} \log y$ is a solution of the corresponding differential equation $(x^{2}+y^{2}) \frac{dy}{dx}-xy=0$.

Form the differential equation of the family of hyperbolas having foci on the $x$-axis and centre at the origin.

The differential equation representing the family of ellipses having foci either on the $x$-axis or on the $y$-axis,centered at the origin,and passing through the point $(0,3)$ is:

The differential equation corresponding to the family of circles given by $(x-a)^2+(y-b)^2=4$,where $a$ and $b$ are parameters,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module