What is the solution for x^'=x(t^2-1)(t^2-9),x(0)=1 ?

The solution for x^'=x(t^2-1)(t^2-9),x(0)=1 is x=e^{(t^5)/5-(10t^3)/3+9t}

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x^'=x(t^2-1)(t^2-9),x(0)=1

x^{^{'}} = x (t^{2} - 1) (t^{2} - 9), x (0) = 1

x = e^{\frac{t ^{5}}{5} - \frac{10 t ^{3}}{3} + 9 t}

Solution steps

x^{'} (t) = x (t^{2} - 1) (t^{2} - 9)

Solve separable ODE:

x = e^{\frac{t ^{5}}{5} - \frac{10 t ^{3}}{3} + 9 t}

x = e^{\frac{t ^{5}}{5} - \frac{10 t ^{3}}{3} + 9 t}

What is the solution for x^'=x(t^2-1)(t^2-9),x(0)=1 ?
The solution for x^'=x(t^2-1)(t^2-9),x(0)=1 is x=e^{(t^5)/5-(10t^3)/3+9t}