Question

If \(y + \frac{d}{dx}(xy) = x(\sin x + \log x)\) then

• A. \(y = \cos x + \frac{2 \sin x}{x} + \frac{2}{x^2} \cos x + \frac{x}{3} \log x - \frac{x}{9} + \frac{c}{x^2}\)
• B. \(y = -\cos x - \frac{2}{x} \sin x + \frac{2}{x^2} \cos x + \frac{x}{3} \log x - \frac{x}{9} + \frac{c}{x^2}\)
• C. \(y = -\cos x + \frac{2}{x} \sin x + \frac{2}{x^2} \cos x + \frac{x}{3} \log x - \frac{x}{9} + \frac{c}{x^2}\)
• D. \(y = \cos x - \frac{2}{x} \sin x + \frac{2}{x^3} \cos x + \frac{x}{3} \log x - \frac{x}{9} + \frac{c}{x^2}\)

Hint:

If the equation contains \(\frac{d}{dx}(xy)\), expand it first. Very often it becomes a standard linear differential equation.

Answer 1

The Correct Option is C

Concept:
This is a first-order linear differential equation. First simplify the derivative term, convert to linear form, find the integrating factor, and then integrate. ip

Step 1: Expand \(\frac{d}{dx}(xy)\).
\[ \frac{d}{dx}(xy)=x\frac{dy}{dx}+y \] So the given equation becomes: \[ y + x\frac{dy}{dx} + y = x(\sin x+\log x) \] \[ x\frac{dy}{dx}+2y=x(\sin x+\log x) \] ip

Step 2: Write it in linear form.
Divide by \(x\): \[ \frac{dy}{dx}+\frac{2}{x}y=\sin x+\log x \] This is of the form: \[ \frac{dy}{dx}+Py=Q \] with \[ P=\frac{2}{x} \] ip

Step 3: Find the integrating factor.
\[ \text{I.F.}=e^{\int \frac{2}{x}\,dx}=e^{2\ln x}=x^2 \] ip

Step 4: Multiply through by the integrating factor.
\[ x^2\frac{dy}{dx}+2xy=x^2\sin x+x^2\log x \] The left side becomes: \[ \frac{d}{dx}(x^2y)=x^2\sin x+x^2\log x \] So, \[ x^2y=\int x^2\sin x\,dx+\int x^2\log x\,dx + c \] ip

Step 5: Integrate the two terms.
Using standard integration by parts: \[ \int x^2\sin x\,dx = -x^2\cos x + 2x\sin x + 2\cos x \] Also, \[ \int x^2\log x\,dx = \frac{x^3}{3}\log x - \frac{x^3}{9} \] Therefore, \[ x^2y= -x^2\cos x + 2x\sin x + 2\cos x + \frac{x^3}{3}\log x - \frac{x^3}{9} + c \] ip

Step 6: Divide by \(x^2\).
\[ y= -\cos x + \frac{2}{x}\sin x + \frac{2}{x^2}\cos x + \frac{x}{3}\log x - \frac{x}{9} + \frac{c}{x^2} \] ip Hence, the correct answer is:
\[ \boxed{(C)} \]