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

The solution for x^'=x^2(t^2-1)(t^2-9),x(0)=1 is x=-(15)/(3t^5-50t^3+135t-15)

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

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

x = - \frac{15}{3 t ^{5} - 50 t ^{3} + 135 t - 15}

Solution steps

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

Solve separable ODE:

x = - \frac{15}{3 t ^{5} - 50 t ^{3} + 135 t - 15}

x = - \frac{15}{3 t ^{5} - 50 t ^{3} + 135 t - 15}

\frac{d y}{d x} + y = y^{- 2}

2 d y - (y - 7)^{2} d x = 0

\frac{d}{d x} (y^{3}) = 50 x

\frac{d y}{d x} = \frac{1}{y - 2}

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

What is the solution for x^'=x^2(t^2-1)(t^2-9),x(0)=1 ?
The solution for x^'=x^2(t^2-1)(t^2-9),x(0)=1 is x=-(15)/(3t^5-50t^3+135t-15)