What is the solution for y^'y^{''}=1 ?

The solution for y^'y^{''}=1 is y= 1/3 (2t+c_{1})^{3/2}+c_{2},y=-1/3 (2t+c_{1})^{3/2}+c_{2}

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y^'y^{''}=1

y^{^{'}} y^{^{''}} = 1

y = \frac{1}{3} (2 t + c_{1})^{\frac{3}{2}} + c_{2}, y = - \frac{1}{3} (2 t + c_{1})^{\frac{3}{2}} + c_{2}

Solution steps

y^{'} (t) y^{''} (t) = 1

Solve using substitution:

y = \frac{1}{3} (2 t + c_{1})^{\frac{3}{2}} + c_{2}, y = - \frac{1}{3} (2 t + c_{1})^{\frac{3}{2}} + c_{2}

y = \frac{1}{3} (2 t + c_{1})^{\frac{3}{2}} + c_{2}, y = - \frac{1}{3} (2 t + c_{1})^{\frac{3}{2}} + c_{2}

What is the solution for y^'y^{''}=1 ?
The solution for y^'y^{''}=1 is y= 1/3 (2t+c_{1})^{3/2}+c_{2},y=-1/3 (2t+c_{1})^{3/2}+c_{2}