The solution of the differential equation sin | vedclass

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

Question

(B) Given the differential equation $\sin \left( \frac{dy}{dx} \right) = a$. Taking the inverse sine on both sides,we get $\frac{dy}{dx} = \sin^{-1}(a

Vedclass · Accepted Answer

$\sin \left( \frac{y - 1}{x} \right) = a$

Vedclass · Answer

(B) Given the differential equation $\sin \left( \frac{dy}{dx} \right) = a$. Taking the inverse sine on both sides,we get $\frac{dy}{dx} = \sin^{-1}(a)$. This is a variable separable differential equation. Integrating both sides with respect to $x$,we have $\int dy = \int \sin^{-1}(a) \, dx$. Since $\sin^{-1}(a)$ is a constant,we get $y = x \sin^{-1}(a) + c$. Using the initial condition $y(0) = 1$,we substitute $x = 0$ and $y = 1$: $1 = 0 \cdot \sin^{-1}(a) + c \implies c = 1$. Thus,the equation becomes $y = x \sin^{-1}(a) + 1$. Rearranging the terms,$y - 1 = x \sin^{-1}(a)$,which implies $\frac{y - 1}{x} = \sin^{-1}(a)$. Taking the sine of both sides,we obtain $\sin \left( \frac{y - 1}{x} \right) = a$.

The solution of the differential equation $\sin \left( \frac{dy}{dx} \right) = a$ with the initial condition $y(0) = 1$ is:

Similar Questions

The equation of the curve whose slope is $\frac{y - 1}{x^2 + x}$ and which passes through the point $(1, 0)$ is

The general solution of the differential equation $\frac{x dy - y dx}{y} = 0$ is . . . . . . .

If the curve $y = y(x)$ passes through the point $(1, e)$ and satisfies the differential equation $dy = y(2 + \log_e x) dx$, $x > 0$, then $y(e)$ is equal to:

The solution of the equation $\frac{dy}{dx} = \frac{1}{x+y+1}$ is

The general solution of the differential equation $\sqrt{1+x^{2}+y^{2}+x^{2} y^{2}}+x y \frac{d y}{d x}=0$ is (where $C$ is a constant of integration)

Vedclass Test Series

Exam Paper Generator

Online Exam Module