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

Find the general solution of $x^4\,y'' - (x^3 + 2x\,y)\,y' + 4y^2 = 0$, using homogeneity to reduce its order.

Step 1: Check homogeneity

Assign weights $[x]=1$, $[y]=k$. Each term must carry the same weight: \[[x^4y''] = k+2,\qquad [x^3y'] = k+2,\qquad [xy\,y'] = 2k,\qquad [y^2]=2k.\]

Setting $k+2=2k$ gives $k=2$: the equation is homogeneous with $[x]=1$, $[y]=2$.

Step 2: Substitution to reduce order

Set $y = v\,x^2$, where $v=v(x)$. Then $y'=v'x^2+2vx$, $y''=v''x^2+4v'x+2v$. Substituting into the equation and simplifying (the $v^2x^4$ terms cancel exactly): $v''x^6 + (3-2v)v'x^5 = 0.$

Divide by $x^5$: \[x\,v'' + (3-2v)\,v' = 0. \tag{1}\]

Step 3: Reduce the order

Let $p = v'$, and treat $p$ as a function of $v$, so $v'' = p\,dp/dx = p\,\dfrac{dp}{dv}\cdot\dfrac{dv}{dx}\cdot\dfrac1p$… more simply $dp/dx = p\,dp/dv$. Equation (1) becomes $x\,p\,dp/dv + (3-2v)p = 0$, and dividing by $p$:

\[ x\,\frac{dp}{dv} = 2v - 3. \tag{2} \]

Also $\dfrac{dx}{dv} = \dfrac1p$. Combine these to compute $\dfrac{d(xp)}{dv} = x\dfrac{dp}{dv} + p\dfrac{dx}{dv} = (2v-3) + 1 = 2(v-1)$: $\frac{d(xp)}{dv} = 2(v-1).$

Integrate: \[xp = (v-1)^2 + C_1,\qquad\text{i.e.}\qquad x\,\frac{dv}{dx} = (v-1)^2 + C_1. \tag{3}\]

Step 4: Solve equation (3)

This is separable: $\dfrac{dv}{(v-1)^2+C_1} = \dfrac{dx}{x}$. The sign of $C_1$ determines the form.

Step 5: General solution

Since $y = vx^2$:

and for $C_1=-a^2<0$: $y = x^2\!\left[1+a\cdot\dfrac{1+C_2^{2a}x^{2a}}{1-C_2^{2a}x^{2a}}\right]$, where $a>0$ and $C_2$ are arbitrary constants.

Step 6: Verification (Case B)

For $y = x^2\!\left(1-\dfrac{1}{L}\right)$, $L=\ln|C_2 x|$: $v' = \dfrac{1}{xL^2}$, $v'' = -\dfrac{1}{x^2L^2}\!\left(1+\dfrac2L\right)$. Check $xv''+(3-2v)v'=0$: \[-\frac{1}{xL^2}\!\left(1+\frac2L\right) + \left(1+\frac2L\right)\frac{1}{xL^2} = 0.\ \checkmark\]