$\left|\begin{array}{ccc}x^2 & y^2 & x+y \\ 2 x & 2 y^{\prime} y & y^{\prime}+1 \\ 2 & 2 y y^{\prime \prime} & y^{\prime \prime}\end{array}\right|=0$

$\left|\begin{array}{ccc}x^2 & y^2 & x+y \\ 2 x & 2 y y^{\prime} & 1+y^{\prime} \\ 2 & 2\left(y^{\prime 2}+y y^{\prime \prime}\right) & y^{\prime \prime}\end{array}\right|=0$

$\left|\begin{array}{ccc}x^2 & y^2 & x+y \\ 2 x & 2 y y^{\prime} & y+1 \\ 2 & 2\left(y^{\prime 2}+y^{\prime} y^{\prime \prime}\right) & y^{\prime \prime}\end{array}\right|=0$

$\left|\begin{array}{ccc}x^2 & y^2 & x+y \\ 2 x & 2 y & 1+y^{\prime} \\ 2 & 2 y y^{\prime} y^{\prime \prime} & y^{\prime \prime}\end{array}\right|=0$

$\left|x+2 y+\frac{3}{5}\right|^{2 / 25} \cdot e^{115(x+2 y)}=C$

$\left|x-2 y+\frac{3}{5}\right|^{2 / 25} \cdot e^{1 / 5(x-2 y)}=C$

$\left|x-2 y+\frac{3}{5}\right|^{\frac{2}{25}} \cdot e^{1 / 5(6 x-2 y)}=C$

$\left|x-2 y+\frac{1}{5}\right|^{\frac{2}{25}} \cdot e^{1 / 5(x-2 y)}=C$

$x=\left(\tan ^{-1} y\right)-1+C e^{-\tan ^{-1} y}$

$x=\left(\tan ^{-1} y\right)-1+C e^{\tan ^{-1} y}$

$x=\left(\tan ^{-1} y\right)-1+C$

$x=\left(\tan ^{-1} y\right)+C e^{-\tan ^{-1} y}$

0

$(a+b)^2$

$a^2-b^2$

$a^2+b^2$