What is the solution for (1+(y^')^2)^{3/2}=y^{''} ?

The solution for (1+(y^')^2)^{3/2}=y^{''} is

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(1+(y^')^2)^{3/2}=y^{''}

(1 + (y^{^{'}})^{2})^{\frac{3}{2}} = y^{^{''}}

y = \int \frac{- t ^{2} - 2 c _{1} t - c _{1}^{2}}{- 1 + c _{1}^{2} + 2 c _{1} t + t ^{2}} d t, y = \int - \frac{- t ^{2} - 2 c _{1} t - c _{1}^{2}}{- 1 + c _{1}^{2} + 2 c _{1} t + t ^{2}} d t

Solution steps

(1 + (y^{'} (t))^{2})^{\frac{3}{2}} = y^{''} (t)

Solve using substitution:

y = \int \frac{- t ^{2} - 2 c _{1} t - c _{1}^{2}}{- 1 + c _{1}^{2} + 2 c _{1} t + t ^{2}} d t, y = \int - \frac{- t ^{2} - 2 c _{1} t - c _{1}^{2}}{- 1 + c _{1}^{2} + 2 c _{1} t + t ^{2}} d t

y = \int \frac{- t ^{2} - 2 c _{1} t - c _{1}^{2}}{- 1 + c _{1}^{2} + 2 c _{1} t + t ^{2}} d t, y = \int - \frac{- t ^{2} - 2 c _{1} t - c _{1}^{2}}{- 1 + c _{1}^{2} + 2 c _{1} t + t ^{2}} d t

\frac{( 2 x + x ^{2} )}{( 1 + x ) ^{2}}

\frac{d y}{d x} = \frac{1}{4 - x ^{2}}

ln (x (1 - x)^{8})

\frac{d ^{2}}{d x ^{2}} (e^{7 x^{2}})

\frac{d ^{2}}{d r ( T ) d T} (r (T) (T)^{2} \frac{d ^{2}}{d r ( T ) d T} (T))

What is the solution for (1+(y^')^2)^{3/2}=y^{''} ?
The solution for (1+(y^')^2)^{3/2}=y^{''} is