What is the solution for y^'=t^{0.5},y(0)=0 ?

The solution for y^'=t^{0.5},y(0)=0 is y=0.66666…t^{1.5}

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y^'=t^{0.5},y(0)=0

y^{^{'}} = t^{0.5}, y (0) = 0

y = 0.66666 \dots t^{1.5}

Solution steps

y^{'} (t) = t^{0.5}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int t^{0.5} d t

\int t^{0.5} d t = 0.66666 \dots t^{1.5} + c_{1}

y = 0.66666 \dots t^{1.5} + c_{1}

Apply initial conditions:

y = 0.66666 \dots t^{1.5}

y = 0.66666 \dots t^{1.5}

\frac{d}{d x} (5 x e^{3 x})

\int_{0}^{\frac{π}{3}} 8 sec^{3} (x) d x

f (x) = 7 x + 2, (2, 4)

x y^{^{'}} + y = e^{x}, y (1) = 1

(3, 6), (5, - 8)

What is the solution for y^'=t^{0.5},y(0)=0 ?
The solution for y^'=t^{0.5},y(0)=0 is y=0.66666…t^{1.5}