What is the solution for t^3(dy)/(dt)+3t^2y=2cos(t) ?

The solution for t^3(dy)/(dt)+3t^2y=2cos(t) is y=(2sin(t))/(t^3)+(c_{1})/(t^3)

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t^3(dy)/(dt)+3t^2y=2cos(t)

t^{3} \frac{d y}{d t} + 3 t^{2} y = 2 cos (t)

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

Solution steps

t^{3} \frac{d y}{d t} + 3 t^{2} y = 2 cos (t)

Solve linear ODE:

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

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

\int x^{15} e^{- x^{16}} d x

f (t) = 2 t^{3} - 3 t^{2} - 4 t

\frac{\partial}{\partial x} (\frac{4 π ^{2} x}{y ^{2}})

9 x e^{x} csc (x)

\frac{1}{x}, 1, 2

What is the solution for t^3(dy)/(dt)+3t^2y=2cos(t) ?
The solution for t^3(dy)/(dt)+3t^2y=2cos(t) is y=(2sin(t))/(t^3)+(c_{1})/(t^3)