Question

Find the general solution of the differential equation \(\dfrac{dy}{dx} + y\cot x = \csc x\).

• A. \(y\sin x = x + c\)
• B. \(y\cos x = x + c\)
• C. \(y\sin x = c - x\)
• D. \(y\cos x = c + x\)

Hint:

For linear differential equations \( \frac{dy}{dx}+Py=Q \), first compute the integrating factor \(e^{\int Pdx}\). Multiplying the equation by it converts the left side into an exact derivative.

Answer 1

The Correct Option is A

Concept: The given differential equation is a linear differential equation of the form \[ \frac{dy}{dx} + Py = Q \] where \(P = \cot x\) and \(Q = \csc x\). The integrating factor (I.F.) is: \[ I.F. = e^{\int P\,dx} \]

Step 1: Find the integrating factor. \[ I.F. = e^{\int \cot x\,dx} \] \[ = e^{\ln(\sin x)} \] \[ = \sin x \]

Step 2: Multiply the equation by the integrating factor. \[ \sin x \frac{dy}{dx} + y\sin x \cot x = \sin x \csc x \] Since \[ \sin x \cot x = \cos x \] the equation becomes \[ \sin x \frac{dy}{dx} + y\cos x = 1 \]

Step 3: Recognize the derivative form. \[ \frac{d}{dx}(y\sin x) = 1 \]

Step 4: Integrate both sides. \[ y\sin x = \int 1\,dx \] \[ y\sin x = x + c \] Thus, the general solution is \[ \boxed{y\sin x = x + c} \]