What is the solution for (dx}{dt}=\frac{t+sin^2(t))/x ,x(0)=1 ?

The solution for (dx}{dt}=\frac{t+sin^2(t))/x ,x(0)=1 is x=sqrt((2t^2+2t-sin(2t)+2)/2)

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(dx}{dt}=\frac{t+sin^2(t))/x ,x(0)=1

\frac{d x}{d t} = \frac{t + sin ^{2} ( t )}{x}, x (0) = 1

x = \frac{2 t ^{2} + 2 t - sin ( 2 t ) + 2}{2}

Solution steps

\frac{d x}{d t} = \frac{t + sin ^{2} ( t )}{x}

Solve separable ODE:

x = \frac{2 t ^{2} + 2 t - sin ( 2 t ) + 2}{2}

x = \frac{2 t ^{2} + 2 t - sin ( 2 t ) + 2}{2}

\frac{d y}{d x} = 0.9 (y - 300)

\frac{d y}{1} = \frac{d x}{1 + y}

\frac{d x}{d t} = - k x, x (0) = 5

\frac{d y}{d x} = (2 y + 1)^{2} x e^{- x}

\frac{d x}{d t} = - 300 \cdot \frac{x}{1 0 ^{8}}

What is the solution for (dx}{dt}=\frac{t+sin^2(t))/x ,x(0)=1 ?
The solution for (dx}{dt}=\frac{t+sin^2(t))/x ,x(0)=1 is x=sqrt((2t^2+2t-sin(2t)+2)/2)