What is the solution for y^'=y(5t^4-6t^5),y(0)=1 ?

The solution for y^'=y(5t^4-6t^5),y(0)=1 is y=e^{-t^6+t^5}

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y^'=y(5t^4-6t^5),y(0)=1

y^{^{'}} = y (5 t^{4} - 6 t^{5}), y (0) = 1

y = e^{- t^{6} + t^{5}}

Solution steps

y^{'} (t) = y (5 t^{4} - 6 t^{5})

Solve separable ODE:

y = e^{- t^{6} + t^{5}}

y = e^{- t^{6} + t^{5}}

y^{^{'}} - k^{2} y = 0

y^{^{'}} - 2 y^{\frac{1}{3}} = 0, y (0) = 0

y^{^{'}} - \frac{7 y}{5160} = 0.07

\frac{d y}{d x} = \frac{x cos ( x )}{y}, y (0) = - 2

e^{y} y^{^{'}} = \frac{( ( ln ( t ) ) ^{2} )}{t}

What is the solution for y^'=y(5t^4-6t^5),y(0)=1 ?
The solution for y^'=y(5t^4-6t^5),y(0)=1 is y=e^{-t^6+t^5}