Find the general solution of x y y'' + y y' + x^2(y')^3 = 0…

Find the general solution of $x\,y\,y'' + y\,y' + x^2(y')^3 = 0$, using homogeneity to reduce its order.

Step 1: Check homogeneity

Assign weights $[x]=1$, $[y]=k$. Each term has weight $[xyy'']=2k-1$, $[yy']=2k-1$, $[x^2(y')^3]=3k-1$. Setting $2k-1=3k-1$ gives $k=0$: $y$ is dimensionless.

Step 2: Substitution motivated by homogeneity

Since $[y]=0$, $[x]=1$, set:

\[ y' = \frac{w}{x}, \tag{1} \] where $w=w(x)$. Then $y'' = \dfrac{w'}{x}-\dfrac{w}{x^2}$. Substitute into the equation: \[xy\!\left(\frac{w'}{x}-\frac{w}{x^2}\right) + y\cdot\frac{w}{x} + x^2\cdot\frac{w^3}{x^3} = 0 \implies y\,w' - \frac{yw}{x} + \frac{yw}{x} + \frac{w^3}{x} = 0.\]

The middle terms cancel: \[y\,w' + \frac{w^3}{x} = 0. \tag{2}\]

Step 3: Eliminate $x$

Since $dy = \dfrac{w}{x}\,dx$, we have $dx = \dfrac{x\,dy}{w}$. Substitute into (2), written as $y\,dw = -\dfrac{w^3}{x}\,dx$: $y\,dw = -\frac{w^3}{x}\cdot\frac{x\,dy}{w} = -w^2\,dy.$

Thus: \[\frac{dw}{w^2} = -\frac{dy}{y}. \tag{3}\]

Step 4: Integrate (3)

\[ -\frac{1}{w} = -\ln|y| + \ln|C_1| = \ln\left|\frac{C_1}{y}\right|,\qquad w = \frac{1}{\ln(y/C_1)}. \tag{4} \]

Step 5: Second integration

Recall $w = xy'$, so $y' = \dfrac{1}{x\ln(y/C_1)}$. Separate variables: $\ln\!\left(\frac{y}{C_1}\right)dy = \frac{dx}{x}.$

The left side integrates by parts ($\int \ln(y/C_1)\,dy = y\ln(y/C_1) - y$): $y\!\left(\ln\frac{y}{C_1} - 1\right) = \ln|C_2\,x|.$

Step 6: General solution

where $C_1 > 0$ and $C_2 \neq 0$ are arbitrary constants.

Step 7: Verification

Implicit differentiation of the general solution reproduces $\ln(y/C_1)\,dy = dx/x$, i.e. $y' = \dfrac{1}{x\ln(y/C_1)}$, confirming step 5. Differentiating once more: $y'' = -\frac{y'}{x} - \frac{(y')^2}{y\ln(y/C_1)}.$

Substitute into $xyy''+yy'+x^2(y')^3$; using $\dfrac{1}{\ln(y/C_1)}=xy'$: $-yy' - \frac{x(y')^2}{\ln(y/C_1)} + yy' + x^2(y')^3 = -x^2(y')^3 + x^2(y')^3 = 0.$ $\checkmark$