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}

z \to 0 lim (\frac{1}{1 + z ^{2}})

\frac{d}{d x} (\frac{ln ( x ^{2} )}{x})

(t^{2} + 1) f^{^{'}} (t) + [f (t)]^{2} + 1 = 0

\int_{2}^{3} \frac{6}{3 - x} d x

\frac{d}{d x} ((\frac{x + 8}{x - 8})^{5})

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