Question

dy/dx=y/x+sin y/x

Answer 1

$$\frac {d y (x)}{d x} = \sin \left(\frac {y (x)}{x}\right) + \frac {y (x)}{x}$$

Let $y(x) = x \, v(x)$, which gives $\frac{dy(x)}{dx} = x \, \frac{dv(x)}{dx} + v(x)$:

$$x \, \frac {dv(x)}{dx} + v(x) = \sin(v(x)) + v(x)$$

Solve for $\frac{dv(x)}{dx}$:

$$\frac {dv(x)}{dx} = \frac {\sin(v(x))}{x}$$

Divide both sides by $\sin(v(x))$:

$$\csc(v(x)) \frac {dv(x)}{dx} = \frac {1}{x}$$

Integrate both sides with respect to $x$:

$$\int \csc(v(x)) \frac {dv(x)}{dx} \, dx = \int \frac {1}{x} \, dx$$

Evaluate the integrals:

$$- \log \left(\cos \left(\frac {v (x)}{2}\right)\right) + \log \left(\sin \left(\frac {v (x)}{2}\right)\right) =$$ $$\log(x) + c_1, \text{ where } c_1 \text{ is an arbitrary constant.}$$

Solve for $v(x)$:

$$v(x) = 2 \cot^{-1} \left(\frac {c - c_1}{x}\right)$$

Simplify the arbitrary constant:

$$v(x) = 2 \cot^{-1} \left(\frac {c_1}{x}\right)$$

Substitute back for $y(x) = x \, v(x)$:

$$y(x) = 2 x \cot^{-1} \left(\frac {c_1}{x}\right)$$