Find the general solution of x y'' = y' + xsin(y')/(x)

Find the general solution of $x\,y'' = y' + x\sin\frac{y'}{x}$.

Step 1: First substitution

Let $p = y'$, so $p' = y''$. The equation becomes: \[x\,p' = p + x\sin\frac{p}{x}. \tag{1}\]

Step 2: Second substitution

This is a homogeneous ODE in $p(x)$. Set $v = \dfrac{p}{x}$, so $p = vx$ and $p' = v + xv'$. Substitute into (1): \[x(v + x\,v') = vx + x\sin v,\qquad xv + x^2 v' = vx + x\sin v,\qquad x^2 v' = x\sin v.\]

Thus: \[x\,v' = \sin v. \tag{2}\]

Step 3: Separate variables

\[ \frac{dv}{\sin v} = \frac{dx}{x}. \]

Step 4: Integrate both sides

\[ \int \csc v\,dv = \int \frac{dx}{x},\qquad \ln\left|\tan\frac{v}{2}\right| = \ln|x| + \ln|C_1|,\qquad \tan\frac{v}{2} = C_1 x. \]

Substitute back $v = \dfrac{p}{x} = \dfrac{y'}{x}$: $\tan\frac{y'}{2x} = C_1 x.$

Solve for $y'$: \[y' = 2x\arctan(C_1 x). \tag{3}\]

Step 5: Second integration

Integrate (3) by parts with $u = \arctan(C_1 x)$, $dw = 2x\,dx$, so $du = \dfrac{C_1}{1+C_1^2 x^2}\,dx$, $w = x^2$: $y = x^2\arctan(C_1 x) - \int \frac{C_1 x^2}{1+C_1^2 x^2}\,dx + C_2.$

For the remaining integral, write $\dfrac{C_1 x^2}{1+C_1^2 x^2} = \dfrac{1}{C_1}\left(1 - \dfrac{1}{1+C_1^2 x^2}\right)$: \[\int \frac{C_1 x^2}{1+C_1^2 x^2}\,dx = \frac{x}{C_1} - \frac{1}{C_1^2}\arctan(C_1 x).\]

Step 6: General solution

\[ y = x^2\arctan(C_1 x) - \frac{x}{C_1} + \frac{1}{C_1^2}\arctan(C_1 x) + C_2, \]

Step 7: Verification

Differentiate the general solution: \[y' = 2x\arctan(C_1 x) + \left(x^2+\frac{1}{C_1^2}\right)\frac{C_1}{1+C_1^2 x^2} - \frac{1}{C_1}.\]

The middle term simplifies to $\dfrac{1}{C_1}$, so $y' = 2x\arctan(C_1 x)$, confirming (3). Differentiate again: $y'' = 2\arctan(C_1 x) + \frac{2C_1 x}{1+C_1^2 x^2}.$

Since $v = \dfrac{y'}{x} = 2\arctan(C_1 x)$, we have $\tan\dfrac{v}{2} = C_1 x$, so $\sin v = \dfrac{2C_1 x}{1+C_1^2 x^2}$. Thus: \[xy'' - y' - x\sin\frac{y'}{x} = 2x\arctan(C_1 x) + \frac{2C_1 x^2}{1+C_1^2 x^2} - 2x\arctan(C_1 x) - \frac{2C_1 x^2}{1+C_1^2 x^2} = 0.\ \checkmark\]