The particular solution of the differential e | vedclass

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

Question

(A) Given the differential equation: $(y + x \frac{dy}{dx}) \sin(xy) = \cos x$ Let $u = xy$. Then,differentiating with respect to $x$,we get $\frac{du

Vedclass · Accepted Answer

$\sin x + \cos(xy) = 1$

Vedclass · Answer

(A) Given the differential equation: $(y + x \frac{dy}{dx}) \sin(xy) = \cos x$ Let $u = xy$. Then,differentiating with respect to $x$,we get $\frac{du}{dx} = y + x \frac{dy}{dx}$. Substituting this into the equation,we have: $\sin(u) \frac{du}{dx} = \cos x$. Integrating both sides with respect to $x$: $\int \sin(u) du = \int \cos x dx$. This yields: $-\cos(u) = \sin x + C$. Substituting $u = xy$ back,we get: $-\cos(xy) = \sin x + C$. At $x = 0$,the value of $y$ is not explicitly given,but since $\cos(xy) = -\sin x - C$,at $x=0$,$\cos(0) = -\sin(0) - C$,which implies $1 = 0 - C$,so $C = -1$. Thus,$-\cos(xy) = \sin x - 1$,which simplifies to $\sin x + \cos(xy) = 1$.

The particular solution of the differential equation $(y + x \cdot \frac{dy}{dx}) \cdot \sin(xy) = \cos x$ at $x = 0$ is

Similar Questions

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

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

The solution of $\log \left(\frac{dy}{dx}\right) = 2x - 5y$ with the initial condition $y(0) = 0$ is:

If $dx + dy = (x + y)(dx - dy)$, then $\log(x + y)$ is equal to

The solution of the differential equation $(x^2 + 1) \frac{dy}{dx} + (y^2 + 1) = 0$ is . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module