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

The solution for y^'=-(3y)/t-2-t^{-4},y(1)=4 is y=-t/2-(ln(t))/(t^3)+9/(2t^3)

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y^'=-(3y)/t-2-t^{-4},y(1)=4

y^{^{'}} = - \frac{3 y}{t} - 2 - t^{- 4}, y (1) = 4

y = - \frac{t}{2} - \frac{ln ( t )}{t ^{3}} + \frac{9}{2 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{9}{2 t ^{3}}

y = - \frac{t}{2} - \frac{ln ( t )}{t ^{3}} + \frac{9}{2 t ^{3}}

\frac{\partial}{\partial y} (\frac{x - 3}{y ^{2} + 1})

x \to - \infty lim (x - 5)

\int_{- 2}^{2} 32 - 8 y^{2} d y

\int_{6}^{8} \frac{46}{( x - 6 ) ^{3}} d x

f (x) = \frac{6 x}{x ^{2} - 3}, (3, 3)

What is the solution for y^'=-(3y)/t-2-t^{-4},y(1)=4 ?
The solution for y^'=-(3y)/t-2-t^{-4},y(1)=4 is y=-t/2-(ln(t))/(t^3)+9/(2t^3)