The solution of the differential equation fra | vedclass

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

Question

(A) Let $v = x + y$. Then $\frac{dv}{dx} = 1 + \frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{dv}{dx} - 1$.Substituting into the equation: $\frac{dv}{dx} -

Vedclass · Accepted Answer

$\log |1 + 	an (\frac{x + y}{2})| = x + c$, where $c$ is the constant of integration

Vedclass · Answer

(A) Let $v = x + y$. Then $\frac{dv}{dx} = 1 + \frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{dv}{dx} - 1$.Substituting into the equation: $\frac{dv}{dx} - 1 = \sin v + \cos v$.$\frac{dv}{dx} = 1 + \sin v + \cos v$.Using half-angle formulas: $1 + \sin v + \cos v = 2\cos^2(\frac{v}{2}) + 2\sin(\frac{v}{2})\cos(\frac{v}{2}) = 2\cos^2(\frac{v}{2}) [1 + 	an(\frac{v}{2})]$.So, $\int \frac{dv}{2\cos^2(\frac{v}{2}) [1 + 	an(\frac{v}{2})]} = \int dx$.Let $u = 1 + 	an(\frac{v}{2})$, then $du = \frac{1}{2} \sec^2(\frac{v}{2}) dv = \frac{1}{2\cos^2(\frac{v}{2})} dv$.Thus, $\int \frac{du}{u} = \int dx \implies \log |u| = x + c$.Substituting back: $\log |1 + 	an(\frac{x + y}{2})| = x + c$.

The solution of the differential equation $\frac{dy}{dx} = \sin(x + y) + \cos(x + y)$ is

Similar Questions

Let $y=y(x)$ be a solution curve of the differential equation,$(1-x^2 y^2) dx = y dx + x dy$. If the line $x = 1$ intersects the curve $y = y(x)$ at $y = 2$ and the line $x = 2$ intersects the curve $y = y(x)$ at $y = \alpha$,then a value of $\alpha$ is

The solution of the differential equation $\frac{dy}{dx} = 2xy$ is

If the solution of $\frac{dy}{dx} - y \log_{e} 0.5 = 0$,$y(0) = 1$,and $y(x) \rightarrow k$,as $x \rightarrow \infty$ then $k =$

Let a solution $y=y(x)$ of the differential equation $x \sqrt{x^2-1} dy - y \sqrt{y^2-1} dx = 0$ satisfy $y(2) = \frac{2}{\sqrt{3}}$.
$STATEMENT-1$: $y(x) = \sec \left(\sec^{-1} x - \frac{\pi}{6}\right)$
$STATEMENT-2$: $y(x)$ is given by $\frac{1}{y} = \frac{2\sqrt{3}}{x} - \sqrt{1 - \frac{1}{x^2}}$

The solution of the differential equation $\frac{dy}{dx} = (ae^{bx} + c\cos mx)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module