What is the solution for (dx)/(dt)=1.5tcos^2(x),x(0)=-1.4 ?

The solution for (dx)/(dt)=1.5tcos^2(x),x(0)=-1.4 is x=arctan(0.75t^2-tan(7/5))+pin

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(dx)/(dt)=1.5tcos^2(x),x(0)=-1.4

\frac{d x}{d t} = 1.5 t cos^{2} (x), x (0) = - 1.4

x = arctan (0.75 t^{2} - tan (\frac{7}{5})) + πn

Solution steps

\frac{d x}{d t} = 1.5 t cos^{2} (x)

Solve separable ODE:

x = arctan (0.75 t^{2} - tan (\frac{7}{5})) + πn

x = arctan (0.75 t^{2} - tan (\frac{7}{5})) + πn

f (x) = 4 x - 4 x

n = 2 \sum \infty \frac{7}{6 ^{n}}

\int 23 arctan (x) d x

f (x) = cos (5 x^{2} + 5)

\frac{d y}{d x} = y x^{5} sin (x^{6})

What is the solution for (dx)/(dt)=1.5tcos^2(x),x(0)=-1.4 ?
The solution for (dx)/(dt)=1.5tcos^2(x),x(0)=-1.4 is x=arctan(0.75t^2-tan(7/5))+pin