Statement (I): The elimination of arbitrary c | vedclass

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

Question

(D) Statement $(I)$: Given $y=(\alpha+\beta+\gamma) x$. Let $k = \alpha+\beta+\gamma$,where $k$ is a single arbitrary constant. Then $y = kx$. Differe

Vedclass · Accepted Answer

$I$ is false and $II$ is true

Vedclass · Answer

(D) Statement $(I)$: Given $y=(\alpha+\beta+\gamma) x$. Let $k = \alpha+\beta+\gamma$,where $k$ is a single arbitrary constant. Then $y = kx$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = k$. Since there is only one independent arbitrary constant,the order of the differential equation is $1$. Thus,Statement $(I)$ is false. Statement $(II)$: Given $y = \alpha x + \beta \sin x + \gamma e^x$. This equation contains $3$ independent arbitrary constants $(\alpha, \beta, \gamma)$. Differentiating with respect to $x$ three times: $(1) \frac{dy}{dx} = \alpha + \beta \cos x + \gamma e^x$ $(2) \frac{d^2y}{dx^2} = -\beta \sin x + \gamma e^x$ $(3) \frac{d^3y}{dx^3} = -\beta \cos x + \gamma e^x$ Since we have $3$ independent arbitrary constants,we can eliminate them to form a differential equation of order $3$. Thus,Statement $(II)$ is true.

Similar Questions

The differential equation of the family of circles passing through the origin and having their centres on the $x$-axis is

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

If the differential equation having $y=Ae^x+B \sin x$ as its general solution is $f(x) \frac{d^2 y}{d x^2}+g(x) \frac{d y}{d x}+h(x) y=0$, then $f(x)+g(x)+h(x)=$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module