The solution of the differential equation fra | vedclass

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

Question

(B) Let $x + y = v$. Differentiating with respect to $x$, we get $1 + \frac{dy}{dx} = \frac{dv}{dx}$, so $\frac{dy}{dx} = \frac{dv}{dx} - 1$.Substitut

Vedclass · Accepted Answer

$	an (\frac{x + y}{2}) = x + c$

Vedclass · Answer

(B) Let $x + y = v$. Differentiating with respect to $x$, we get $1 + \frac{dy}{dx} = \frac{dv}{dx}$, so $\frac{dy}{dx} = \frac{dv}{dx} - 1$.Substituting into the equation: $\frac{dv}{dx} - 1 = \cos v$.$\frac{dv}{dx} = 1 + \cos v = 2 \cos^2 (\frac{v}{2})$.Separating variables: $\frac{dv}{2 \cos^2 (\frac{v}{2})} = dx$.$\frac{1}{2} \sec^2 (\frac{v}{2}) dv = dx$.Integrating both sides: $\int \frac{1}{2} \sec^2 (\frac{v}{2}) dv = \int dx$.$	an (\frac{v}{2}) = x + c$.Substituting $v = x + y$ back: $	an (\frac{x + y}{2}) = x + c$.

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

Similar Questions

The solution of the differential equation $e^{-x}(y + 1)dy + (\cos^2 x - \sin 2x)y dx = 0$, given that $y = 1$ when $x = 0$ is

The solution of the differential equation $e^{-x}(y + 1) dy + (\cos^2 x - \sin 2x)y dx = 0$, given that $y = 1$ when $x = 0$ is

The particular solution of $\frac{dy}{dx} = 1 + x + y^2 + xy^2$,when $y(0) = 0$,is

Find a particular solution satisfying the given condition:
$(x^{3}+x^{2}+x+1) \frac{dy}{dx} = 2x^{2}+x; y=1$ when $x=0$

If the differential equation $\frac{dy}{dx} + \frac{x}{y} = \frac{a}{y}$,where $a$ is a constant,represents a family of circles,then the radius of the circle is ......

Vedclass Test Series

Exam Paper Generator

Online Exam Module