What is the solution for (dx)/(dt)= 3/2 tcos^2(x),x(0)=-1.4 ?

The solution for (dx)/(dt)= 3/2 tcos^2(x),x(0)=-1.4 is x=arctan((3t^2)/4-tan(7/5))+pin

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(dx)/(dt)= 3/2 tcos^2(x),x(0)=-1.4

\frac{d x}{d t} = \frac{3}{2} t cos^{2} (x), x (0) = - 1.4

x = arctan (\frac{3 t ^{2}}{4} - tan (\frac{7}{5})) + πn

Solution steps

\frac{d x}{d t} = \frac{3}{2} t cos^{2} (x)

Solve separable ODE:

x = arctan (\frac{3 t ^{2}}{4} - tan (\frac{7}{5})) + πn

x = arctan (\frac{3 t ^{2}}{4} - tan (\frac{7}{5})) + πn

\frac{d}{d x} (x + 1 + x)

\int 10 sin^{3} (x) cos^{5} (x) d x

\frac{1}{1 - t}

\int \frac{x}{x ^{2} ( x ^{2} + 1 )} d x

y^{^{''}} + 6 y^{^{'}} + 11 y = 0, y (0) = - 1, y^{^{'}} (0) = - 4

What is the solution for (dx)/(dt)= 3/2 tcos^2(x),x(0)=-1.4 ?
The solution for (dx)/(dt)= 3/2 tcos^2(x),x(0)=-1.4 is x=arctan((3t^2)/4-tan(7/5))+pin