If frac{dy}{dx} + 2x tan(x-y) = 1, then sin(x | vedclass

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

Question

(C) Given differential equation is $\frac{dy}{dx} + 2x 	an(x-y) = 1$. Let $t = x - y$. Then $\frac{dt}{dx} = 1 - \frac{dy}{dx}$, which implies $\frac

Vedclass · Accepted Answer

$A e^{x^2}$

Vedclass · Answer

(C) Given differential equation is $\frac{dy}{dx} + 2x 	an(x-y) = 1$. Let $t = x - y$. Then $\frac{dt}{dx} = 1 - \frac{dy}{dx}$, which implies $\frac{dy}{dx} = 1 - \frac{dt}{dx}$. Substituting this into the given equation: $1 - \frac{dt}{dx} + 2x 	an(t) = 1$ $\Rightarrow \frac{dt}{dx} = 2x 	an(t)$ $\Rightarrow \cot(t) dt = 2x dx$. Integrating both sides: $\int \cot(t) dt = \int 2x dx$ $\ln|\sin(t)| = x^2 + C$. Let $C = \ln|A|$, then $\ln|\sin(t)| = x^2 + \ln|A|$. $\ln|\sin(t)| - \ln|A| = x^2$ $\ln|\frac{\sin(t)}{A}| = x^2$ $\sin(t) = A e^{x^2}$. Substituting $t = x - y$ back, we get $\sin(x-y) = A e^{x^2}$.

If $\frac{dy}{dx} + 2x \tan(x-y) = 1$, then $\sin(x-y)$ is equal to

Similar Questions

The general solution of the differential equation $\frac{dy}{dx} = (9x + y + 5)^2$ is:

The solution of the differential equation $\cos y \log(\sec x + \tan x) dx = \cos x \log(\sec y + \tan y) dy$ is

The equation of the curve passing through $(1, 0)$ and having a slope of $\frac{y - 1}{x^2 + x}$ is:

The particular solution of the differential equation $\log\left(\frac{dy}{dx}\right) = x$,when $x = 0, y = 1$ is .....

The solution of the differential equation $dy - \sin x \sin y \, dx = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module