Find the general solution of x^4l((y')^2 - 2y y''r) = 4x^3 y…

Find the general solution of $x^4\bigl((y')^2 - 2y\,y''\bigr) = 4x^3\,y\,y' + 1$.

Step 1: Substitution $v = xy$

Set $v=xy$, so $y=v/x$, $y'=(v'x-v)/x^2$, $y''=(v''x^2-2v'x+2v)/x^3$. Substituting and simplifying (all terms combine into a single perfect-square structure): \[(xv'-v)^2 - 2x^2\,v\,v'' = 1. \tag{1}\]

Step 2: Substitution $A = xv' - v$

Since $A = xv'-v$, differentiating gives $dA/dx = xv''$, so $x^2vv'' = xv\cdot(dA/dx)$. Equation (1) becomes: \[A^2 - 2xv\,\frac{dA}{dx} = 1. \tag{2}\]

Also, from $A=xv'-v$: $v' = (A+v)/x$.

Step 3: Change of variable to $v$

Treat $A$ as a function of $v$: $\dfrac{dA}{dv} = \dfrac{dA/dx}{dv/dx} = \dfrac{(A^2-1)/(2xv)}{(A+v)/x} = \dfrac{A^2-1}{2v(A+v)}$ (the factor of $x$ cancels). Let $w = A+v$:

\[ \frac{dw}{dv} = \frac{dA}{dv}+1 = \frac{A^2-1+2v(A+v)}{2v(A+v)} = \frac{w^2+v^2-1}{2vw}. \tag{3} \]

Step 4: Linearize via $\Phi = w^2$

Let $\Phi=w^2$, so $\Phi' = 2w\,dw/dv$. Equation (3) becomes: \[v\,\frac{d\Phi}{dv} - \Phi = v^2-1,\qquad \frac{d\Phi}{dv} - \frac{\Phi}{v} = v - \frac{1}{v}. \tag{4}\]

Step 5: Solve the linear ODE (4)

Integrating factor $1/v$: $\dfrac{d}{dv}\!\left(\dfrac{\Phi}{v}\right) = 1-\dfrac{1}{v^2}$. Integrate: $\frac{\Phi}{v} = v+\frac{1}{v}+C_1,\qquad \Phi = v^2+C_1v+1.$

Since $\Phi=w^2=(A+v)^2$: \[A+v = \pm\sqrt{v^2+C_1v+1}. \tag{5}\]

Step 6: Find $x(v)$

Since $dx/dv = x/(A+v) = \pm x/\sqrt{v^2+C_1v+1}$: $\frac{dx}{x} = \pm\frac{dv}{\sqrt{v^2+C_1v+1}}.$

Integrate using $\displaystyle\int\frac{dv}{\sqrt{v^2+C_1v+1}} = \ln\!\left(v+\tfrac{C_1}{2}+\sqrt{v^2+C_1v+1}\right)$: $x = C_2\!\left(v+\frac{C_1}{2}+\sqrt{v^2+C_1v+1}\right).$

Step 7: General solution

Substituting back $v=xy$, the general solution is given implicitly: where $C_1$ and $C_2$ are arbitrary constants.

Step 8: Verification

Let $S = \sqrt{x^2y^2+C_1xy+1}$, so $F \equiv xy+\dfrac{C_1}{2}+S-C_2x=0$. Differentiating $F=0$ implicitly and using $S=C_2x-xy-\dfrac{C_1}{2}$, the algebra reduces (as in step 6) to:

\[ y + xy' = \frac{S}{x},\qquad\text{i.e.}\qquad xy+x^2y' = S. \]

Squaring and using $S^2=x^2y^2+C_1xy+1$ recovers $(xv'-v)^2-2x^2vv''=1$ with $v=xy$, which is exactly equation (1) — equivalent to the original equation. $\checkmark$