Question

The solution of the differential equation log x $\frac{d y}{d x}+\frac{y}{x} =$ sin 2x is

• A. $y log \left|x\right| =C-\frac{1}{2} cos\, x$
• B. $y log \left|x\right| =C+\frac{1}{2} cos\, 2 x$
• C. $y log \left|x\right| =C-\frac{1}{2} cos\, 2 x$
• D. $xy log \left|x\right| =C-\frac{1}{2} cos\, 2 x$

Answer 1

Answer: (C)

$y log \left|x\right| =C-\frac{1}{2} cos\, 2 x$

Answer 2

$\frac{d y}{d x}+\frac{y}{x \, log\, x }=\frac{sin \, 2x}{log\, x}$ $I.F. =e^{\int \frac{d x}{x \, log \, x}}$ $\therefore\quad I.F. = e^{\int \frac{1}{t} dt}=e^{log\, t} =t=log \left|x\right|$ solution is given by $y \left(I.F.\right)=\int \left(I.F.\right) dx+C$ $y log \left|x\right|= \int \frac{sin\, 2x}{log\, \left|x\right|} \left(log \left|x\right|\right) dx+C$ $=-\frac{cos\, 2x}{2}+C$