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

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