Find the general solution of y'' = (2x y - (5)/(x))y' + 4y^2…

Find the general solution of $y'' = \left(2x\,y - \frac{5}{x}\right)y' + 4y^2 - \frac{4y}{x^2}$.

Step 1: Substitution

The powers of $x$ suggest $y = v/x^2$. Then: $y' = \frac{v'x-2v}{x^3},\qquad y'' = \frac{v''x^2-4v'x+6v}{x^4}.$

Step 2: Substitute and simplify

Multiply the equation by $x^4$ and substitute; the $v^2$ and constant-times-$v$ terms cancel exactly, leaving: $v''x^2 + v'x - 2v\,v'x = 0.$

Divide by $x$: \[x\,v'' = v'(2v-1). \tag{1}\]

Step 3: Reduce the order

Let $p=v'$, treated as a function of $v$, so $dp/dx = p\,dp/dv$. Equation (1) becomes $xp\,dp/dv = p(2v-1)$, and dividing by $p$: \[x\,\frac{dp}{dv} = 2v-1. \tag{2}\]

Also $dx/dv = 1/p$. Compute $\dfrac{d(xp)}{dv} = \dfrac1p\cdot p + x\cdot(2v-1)/x = 1+(2v-1)=2v$. Integrate: \[x\,p = v^2 + C_1,\qquad\text{i.e.}\qquad x\,\frac{dv}{dx} = v^2+C_1. \tag{3}\]

Step 4: Solve equation (3)

This is separable: $\dfrac{dv}{v^2+C_1} = \dfrac{dx}{x}$. The sign of $C_1$ determines the form.

Step 5: General solution

Since $y = v/x^2$: where $a>0$ and $C_2$ are arbitrary constants.

Step 6: Verification (Case B)

Let $L=\ln|C_2x|$, so $y = -\dfrac{1}{x^2L}$. Direct differentiation gives: \[y' = \frac{1+2L}{L^2x^3},\qquad y'' = \frac{-2L^3-5L^2-6L}{L^3x^4}\cdot\frac{1}{1} = \frac{-2/L^3-5/L^2-6/L}{x^4}.\]

Independently, the right-hand side $\left(2xy-\dfrac5x\right)y'+4y^2-\dfrac{4y}{x^2}$ simplifies to the same expression $\dfrac{-2/L^3-5/L^2-6/L}{x^4}$, confirming $y''$ matches exactly. $\checkmark$