Find the general solution of 5(y''')^2 - 3 y'' y^((4)) = 0…

Find the general solution of $5(y''')^2 - 3\,y''\,y^{(4)} = 0$, by transforming both sides into perfect derivatives.

Step 1: Recognize the perfect derivative

Let $p = y'''/y''$. Since $\dfrac{d}{dx}\!\left(\dfrac{y'''}{y''}\right) = \dfrac{y^{(4)}}{y''} - p^2$, dividing the equation by $(y'')^2$ gives $5p^2-3\left(p'+p^2\right)=0$, i.e.

\[ 3\,\frac{dp}{dx} = 2p^2. \tag{$*$} \]

Step 2: First integration — solve for $p$

Separate variables in ($*$): \[\frac{dp}{p^2} = \frac{2}{3}\,dx \implies \frac{y'''}{y''} = p = \frac{3}{a-2x}.\]

Step 3: Second integration

The equation $y''' = \dfrac{3}{a-2x}\,y''$ is separable: \[\frac{d(y'')}{y''} = \frac{3\,dx}{a-2x} \implies y'' = b\,(a-2x)^{-3/2}.\]

Step 4: Third integration

Integrate $y''=b(a-2x)^{-3/2}$ using $u=a-2x$: $y' = b\,(a-2x)^{-1/2} + C_1.$

Step 5: Fourth integration

Integrate again: $y = -b\sqrt{a-2x} + C_1 x + C_2.$

Relabel $-b\to C_3$, $a\to C_4$:

Step 6: General solution

where $C_1$, $C_2$, $C_3$, $C_4$ are arbitrary constants.

Step 7: Verification

Let $y = C_1x+C_2+C_3\sqrt{C_4-2x}$. Then $y' = C_1-\dfrac{C_3}{\sqrt{C_4-2x}}$, $y''=-\dfrac{C_3}{(C_4-2x)^{3/2}}$, $y'''=-\dfrac{3C_3}{(C_4-2x)^{5/2}}$, $y^{(4)}=-\dfrac{15C_3}{(C_4-2x)^{7/2}}$. Check:

\[ 5(y''')^2 = \frac{45C_3^2}{(C_4-2x)^5},\qquad 3y''y^{(4)} = \frac{45C_3^2}{(C_4-2x)^5}. \]

The two sides are equal, so $5(y''')^2-3y''y^{(4)}=0$. $\checkmark$