What is the solution for y^'=-(3y)/t-2-t^{-4} ?

The solution for y^'=-(3y)/t-2-t^{-4} is y=-t/2-(ln(t))/(t^3)+(c_{1})/(t^3)

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y^'=-(3y)/t-2-t^{-4}

y^{^{'}} = - \frac{3 y}{t} - 2 - t^{- 4}

y = - \frac{t}{2} - \frac{ln ( t )}{t ^{3}} + \frac{c _{1}}{t ^{3}}

Solution steps

y^{'} (t) = - \frac{3 y}{t} - 2 - t^{- 4}

Solve linear ODE:

y = - \frac{t}{2} - \frac{ln ( t )}{t ^{3}} + \frac{c _{1}}{t ^{3}}

y = - \frac{t}{2} - \frac{ln ( t )}{t ^{3}} + \frac{c _{1}}{t ^{3}}

n = 0 \sum \infty \frac{e ^{n}}{3 ^{n}}

\int_{\frac{1}{3}}^{1} \frac{y + 4}{y ^{2} + 2 y} d y

x \to \frac{3 π}{2} lim (ln (1 + sin (x)))

\frac{d}{d y} (2 x + 3 y)

What is the solution for y^'=-(3y)/t-2-t^{-4} ?
The solution for y^'=-(3y)/t-2-t^{-4} is y=-t/2-(ln(t))/(t^3)+(c_{1})/(t^3)