What is the solution for 9t^3(dx)/(dt)+6x-6=0 ?

The solution for 9t^3(dx)/(dt)+6x-6=0 is x=e^{1/(3t^2)-c_{1}}+1

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9t^3(dx)/(dt)+6x-6=0

9 t^{3} \frac{d x}{d t} + 6 x - 6 = 0

x = e^{\frac{1}{3 t ^{2}} - c_{1}} + 1

Solution steps

9 t^{3} \frac{d x}{d t} + 6 x - 6 = 0

Solve separable ODE:

x = e^{\frac{1}{3 t ^{2}} - c_{1}} + 1

x = e^{\frac{1}{3 t ^{2}} - c_{1}} + 1

4 e d y - (y - 1)^{2} d x = 0

9 x^{3} y^{2} \frac{d y}{d x} = - 9 x^{2} y^{3}

\frac{d y}{d x} = \frac{cot ( x )}{y}, y (\frac{π}{2}) = 0

x y^{^{'}} - (1 + x^{2}) y = 0

\frac{d y}{d t} = a y - b

What is the solution for 9t^3(dx)/(dt)+6x-6=0 ?
The solution for 9t^3(dx)/(dt)+6x-6=0 is x=e^{1/(3t^2)-c_{1}}+1