What is the solution for x^'=20t-(4x)/(30)t ?

The solution for x^'=20t-(4x)/(30)t is x=(e^{(-t^2+c_{1})/(15)})/2+150

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x^'=20t-(4x)/(30)t

x^{^{'}} = 20 t - \frac{4 x}{30} t

x = \frac{e ^{\frac{- t ^{2} + c _{1}}{15}}}{2} + 150

Solution steps

x^{'} (t) = 20 t - \frac{4 x}{30} t

Solve separable ODE:

x = \frac{e ^{\frac{- t ^{2} + c _{1}}{15}}}{2} + 150

x = \frac{e ^{\frac{- t ^{2} + c _{1}}{15}}}{2} + 150

2 x y \frac{d y}{d x} = 6 + 14 ln (x)

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

y d x - x d y = y^{3} + y^{2} x d y

\frac{d}{d x} (3 y^{2}) + 2 x^{3} - 14 = 0

x^{^{'}} = x^{3}, x (0) = 1

What is the solution for x^'=20t-(4x)/(30)t ?
The solution for x^'=20t-(4x)/(30)t is x=(e^{(-t^2+c_{1})/(15)})/2+150