What is the solution for t^2y^2+e^{1/t}+(2yt^3+t^2y^3)y^'=0 ?

The solution for t^2y^2+e^{1/t}+(2yt^3+t^2y^3)y^'=0 is

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t^2y^2+e^{1/t}+(2yt^3+t^2y^3)y^'=0

t^{2} y^{2} + e^{\frac{1}{t}} + (2 y t^{3} + t^{2} y^{3}) y^{^{'}} = 0

y = 2^{\frac{1}{2}} (- t + (t^{2} + e^{\frac{1}{t}} + c_{1})^{\frac{1}{2}})^{\frac{1}{2}}, y = - 2^{\frac{1}{2}} (- t + (t^{2} + e^{\frac{1}{t}} + c_{1})^{\frac{1}{2}})^{\frac{1}{2}}, y = 2^{\frac{1}{2}} (- t - (t^{2} + e^{\frac{1}{t}} + c_{1})^{\frac{1}{2}})^{\frac{1}{2}}, y = - 2^{\frac{1}{2}} (- t - (t^{2} + e^{\frac{1}{t}} + c_{1})^{\frac{1}{2}})^{\frac{1}{2}}

Solution steps

t^{2} y^{2} + e^{\frac{1}{t}} + (2 y t^{3} + t^{2} y^{3}) y^{'} (t) = 0

Solve exact ODE:

y = 2^{\frac{1}{2}} (- t + (t^{2} + e^{\frac{1}{t}} + c_{1})^{\frac{1}{2}})^{\frac{1}{2}}, y = - 2^{\frac{1}{2}} (- t + (t^{2} + e^{\frac{1}{t}} + c_{1})^{\frac{1}{2}})^{\frac{1}{2}}, y = 2^{\frac{1}{2}} (- t - (t^{2} + e^{\frac{1}{t}} + c_{1})^{\frac{1}{2}})^{\frac{1}{2}}, y = - 2^{\frac{1}{2}} (- t - (t^{2} + e^{\frac{1}{t}} + c_{1})^{\frac{1}{2}})^{\frac{1}{2}}

y = 2^{\frac{1}{2}} (- t + (t^{2} + e^{\frac{1}{t}} + c_{1})^{\frac{1}{2}})^{\frac{1}{2}}, y = - 2^{\frac{1}{2}} (- t + (t^{2} + e^{\frac{1}{t}} + c_{1})^{\frac{1}{2}})^{\frac{1}{2}}, y = 2^{\frac{1}{2}} (- t - (t^{2} + e^{\frac{1}{t}} + c_{1})^{\frac{1}{2}})^{\frac{1}{2}}, y = - 2^{\frac{1}{2}} (- t - (t^{2} + e^{\frac{1}{t}} + c_{1})^{\frac{1}{2}})^{\frac{1}{2}}

y^{^{''}} + 8 y^{^{'}} + 25 y = 0

y = x^{3} - 2 x + 1, (2, 5)

\int_{- 2}^{1} ∣ x ∣ d x

\frac{d y}{d x} + a y = b

What is the solution for t^2y^2+e^{1/t}+(2yt^3+t^2y^3)y^'=0 ?
The solution for t^2y^2+e^{1/t}+(2yt^3+t^2y^3)y^'=0 is