What is the solution for (dy}{dt}=-\frac{3y)/t-2-t^{-4} ?

The solution for (dy}{dt}=-\frac{3y)/t-2-t^{-4} is y=-t/2-(ln(t))/(t^3)+(c_{1})/(t^3)

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(dy}{dt}=-\frac{3y)/t-2-t^{-4}

\frac{d y}{d t} = - \frac{3 y}{t} - 2 - t^{- 4}

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

Solution steps

\frac{d y}{d 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}}

\frac{\partial}{\partial z} (cos (x y) + sin (yz))

\int sin^{3} (x) cos^{2} (x) d x

\int_{1}^{e^{2}} \frac{ln ^{5} ( x ^{2} )}{x} d x

f (x) = (4 x - 5) (3 x^{3} - x^{2} + 1)

\int \frac{x - 3}{x ^{2} - 1} d x

What is the solution for (dy}{dt}=-\frac{3y)/t-2-t^{-4} ?
The solution for (dy}{dt}=-\frac{3y)/t-2-t^{-4} is y=-t/2-(ln(t))/(t^3)+(c_{1})/(t^3)