What is the solution for y^'=ty^{1/2} ?

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

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y^'=ty^{1/2}

y^{^{'}} = t y^{\frac{1}{2}}

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

Solution steps

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

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}

(t^{2} + 5 t) x^{^{'}} = 5 x + 5, x (1) = 6

y^{^{'}} = \frac{1}{5} y cos^{2} (y)

y^{^{'}} + 2 y - 4 = 0

\frac{d y}{d x} + ln (x) \cdot y = 3 \cdot ln (x)

\frac{d x}{d t} = 14 - (\frac{84}{100}) \cdot x

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