Find the general solution of (x^2+1)l((y')^2 - y y''r) = x y…

Find the general solution of $(x^2+1)\bigl((y')^2 - y\,y''\bigr) = x\,y\,y'$, using homogeneity to reduce its order.

Step 1: Check homogeneity

Assign weights $[x]=1$, $[y]=k$. Every term has weight $2k$, so the equation is homogeneous for any $k$.

Step 2: Logarithmic substitution

Divide by $y^2$ and set $z=\ln y$, so $z'=y'/y$ and $(y')^2-yy'' = -y^2\!\left(\dfrac{yy''-(y')^2}{y^2}\right) = -y^2z''$. The equation becomes: \[(x^2+1)\,z'' + x\,z' = 0. \tag{1}\]

Step 3: Reduce the order

Let $p=z'$, so $p'=z''$. Equation (1) becomes the separable ODE: \[(x^2+1)\,p' + x\,p = 0. \tag{2}\]

Step 4: Solve equation (2)

\[ \frac{dp}{p} = -\frac{x\,dx}{x^2+1} \implies p = \frac{C_1}{\sqrt{x^2+1}}. \]

Thus: \[z' = \frac{C_1}{\sqrt{x^2+1}}. \tag{3}\]

Step 5: Second integration

Using $\displaystyle\int\frac{dx}{\sqrt{x^2+1}} = \ln\!\left(x+\sqrt{x^2+1}\right)$: $z = C_1\ln\!\left(x+\sqrt{x^2+1}\right) + C_2.$

Step 6: General solution

Since $z=\ln y$: where $C_1$ and $C_2$ are arbitrary constants, valid for all $x \in \mathbb{R}$.

Step 7: Verification

Let $\varphi = x+\sqrt{x^2+1}$, so $\varphi' = \varphi/\sqrt{x^2+1}$, and $y = C_2\varphi^{C_1}$. Then $y' = \dfrac{C_1 y}{\sqrt{x^2+1}}$ and $y'' = \dfrac{C_1^2 y}{x^2+1} - \dfrac{C_1 x y}{(x^2+1)^{3/2}}$. Compute:

\[ (y')^2 - yy'' = \frac{C_1 x y^2}{(x^2+1)^{3/2}}. \]

Check $(x^2+1)\bigl((y')^2-yy''\bigr) - xyy'$: $\frac{C_1xy^2}{\sqrt{x^2+1}} - x\cdot\frac{C_1y^2}{\sqrt{x^2+1}} = 0.$ $\checkmark$