Find the general solution of x y'' = y' + xl((y')^2 + x^2r)

Find the general solution of $x\,y'' = y' + x\bigl((y')^2 + x^2\bigr)$.

Step 1: First substitution

Let $p = y'$, so $p' = y''$. The equation becomes: \[x\,p' = p + x(p^2 + x^2),\qquad x\,p' = p + xp^2 + x^3. \tag{1}\]

Divide by $x$: \[p' = \frac{p}{x} + p^2 + x^2. \tag{2}\]

Step 2: Second substitution

Set $v = \dfrac{p}{x}$, so $p = vx$ and $p' = v + xv'$. Substitute into (2): $v + xv' = v + x^2(v^2 + 1),\qquad xv' = x^2(v^2+1).$

Thus: \[v' = x(v^2+1). \tag{3}\]

Step 3: Separate variables

\[ \frac{dv}{v^2+1} = x\,dx. \]

Step 4: Integrate both sides

\[ \int \frac{dv}{v^2+1} = \int x\,dx,\qquad \arctan v = \frac{x^2}{2} + C_1. \]

Thus $v = \tan\!\left(\dfrac{x^2}{2} + C_1\right)$. Substitute back $v = \dfrac{p}{x} = \dfrac{y'}{x}$: \[y' = x\tan\!\left(\frac{x^2}{2} + C_1\right). \tag{4}\]

Step 5: Second integration

Integrate (4) using $u = \dfrac{x^2}{2} + C_1$, $du = x\,dx$: \[y = \int x\tan\!\left(\frac{x^2}{2}+C_1\right)dx = \int \tan u\,du = -\ln|\cos u| + C_2.\]

Step 6: General solution

The solution is recorded in the result below.

Step 7: Verification

Differentiate (write $\theta = \dfrac{x^2}{2}+C_1$ for brevity): $y' = x\tan\theta,\qquad y'' = \tan\theta + \frac{x^2}{\cos^2\theta}.$

Check $xy'' - y' - x(y'^2+x^2)$: \[x\left(\tan\theta + \frac{x^2}{\cos^2\theta}\right) - x\tan\theta - x\left(x^2\tan^2\theta + x^2\right) = x^3\left(\frac{1}{\cos^2\theta} - \tan^2\theta - 1\right).\]

Since $\dfrac{1}{\cos^2\theta} - \tan^2\theta = 1$, the bracket vanishes: $x^3(1-1) = 0.$ $\checkmark$