What is the solution for y^'=tsqrt(y) ?

The solution for y^'=tsqrt(y) is y=(t^4}{16}+\frac{c_{1}t^2)/4+(c_{1}^2)/4

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y^'=tsqrt(y)

y^{^{'}} = t y

y = \frac{t ^{4}}{16} + \frac{c _{1} t ^{2}}{4} + \frac{c _{1}^{2}}{4}

Solution steps

y^{'} (t) = t y

Solve separable ODE:

y = \frac{t ^{4}}{16} + \frac{c _{1} t ^{2}}{4} + \frac{c _{1}^{2}}{4}

y = \frac{t ^{4}}{16} + \frac{c _{1} t ^{2}}{4} + \frac{c _{1}^{2}}{4}

What is the solution for y^'=tsqrt(y) ?
The solution for y^'=tsqrt(y) is y=(t^4}{16}+\frac{c_{1}t^2)/4+(c_{1}^2)/4