The general solution of sin ^{-1}left(frac{d | vedclass

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

Question

(D) Given equation: $\sin ^{-1}\left(\frac{d y}{d x}ight)=x+y$ $	herefore \frac{d y}{d x}=\sin (x+y)$ Let $x+y=t$. Differentiating with respect to

Vedclass · Accepted Answer

$	an (x+y)-\sec (x+y)=x+c$

Vedclass · Answer

(D) Given equation: $\sin ^{-1}\left(\frac{d y}{d x}ight)=x+y$ $	herefore \frac{d y}{d x}=\sin (x+y)$ Let $x+y=t$. Differentiating with respect to $x$,we get $1+\frac{d y}{d x}=\frac{d t}{d x}$,so $\frac{d y}{d x}=\frac{d t}{d x}-1$. Substituting this into the differential equation: $\frac{d t}{d x}-1=\sin t$ $\frac{d t}{d x}=1+\sin t$ Separating variables: $\int \frac{d t}{1+\sin t}=\int d x$ Multiply numerator and denominator by $(1-\sin t)$: $\int \frac{1-\sin t}{1-\sin^2 t} d t=\int d x$ $\int \frac{1-\sin t}{\cos^2 t} d t=\int d x$ $\int (\sec^2 t - \sec t 	an t) d t = \int d x$ Integrating both sides: $	an t - \sec t = x + c$ Substituting $t = x+y$ back: $	an (x+y) - \sec (x+y) = x + c$

The general solution of $\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y$ is

Similar Questions

If $\lim _{x \rightarrow \infty} y(x)=\frac{\pi}{2}$,then the solution of $x^3 \sin y \frac{d y}{d x}=2$ is $\cos y=$

If a curve passes through the point $(1,1)$ and at any point $(x, y)$ on the curve,the product of the slope of its tangent and $x$ coordinate of the point is equal to the $y$ coordinate of the point,then the curve also passes through the point

The general solution of the differential equation $(y \sin x + y) \frac{dy}{dx} - \cos^2 x = 0$ is:

The general solution of $\left(x \frac{dy}{dx} - y\right) \sin \frac{y}{x} = x^3 e^x$ is

The general solution of the differential equation $x+y \frac{dy}{dx}=\sec(x^2+y^2)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module