What is the solution for y^'=3,0.4-0.24y^2y(0)=0 ?

The solution for y^'=3,0.4-0.24y^2y(0)=0 is y=3t+c_{1}

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y^'=3,0.4-0.24y^2y(0)=0

y^{^{'}} = 3, 0.4 - 0.24 y^{2} y (0) = 0

y = 3 t + c_{1}

Solution steps

y^{'} (t) = 3

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

then

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

y = \int 3 d t

\int 3 d t = 3 t + c_{1}

y = 3 t + c_{1}

f^{^{'}} (x) = - sin (x^{5}) - 78 x

y^{^{'}} = 2, x y^{2} y = \frac{1}{( 4 - x ^{2} )}

f^{^{'}} (x) = x^{3} + 8 x - 5 x + 6

y^{^{'}} = \frac{( 2 )}{( x ^{3} ) + 5 x ^{\frac{( 2 )}{( 3 )}}}

x^{2} y^{^{'}} = 1

What is the solution for y^'=3,0.4-0.24y^2y(0)=0 ?
The solution for y^'=3,0.4-0.24y^2y(0)=0 is y=3t+c_{1}