What is the solution for t^3(dy)/(dt)+3t^2y=8cos(t),y(pi)=0 ?

The solution for t^3(dy)/(dt)+3t^2y=8cos(t),y(pi)=0 is y=(8sin(t))/(t^3)

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t^3(dy)/(dt)+3t^2y=8cos(t),y(pi)=0

t^{3} \frac{d y}{d t} + 3 t^{2} y = 8 cos (t), y (π) = 0

y = \frac{8 sin ( t )}{t ^{3}}

Solution steps

t^{3} \frac{d y}{d t} + 3 t^{2} y = 8 cos (t)

Solve linear ODE:

y = \frac{8 sin ( t )}{t ^{3}}

y = \frac{8 sin ( t )}{t ^{3}}

y = (x^{2} - 2 x + 2) e^{\frac{5 x}{2}}

\int sec (θ) tan (θ) d θ

\frac{d}{d t} (p (t) e^{t})

y = 5 + 1 - x, (- 8, 8)

\int_{0}^{12} 2 π y (144 - y^{2}) d y

What is the solution for t^3(dy)/(dt)+3t^2y=8cos(t),y(pi)=0 ?
The solution for t^3(dy)/(dt)+3t^2y=8cos(t),y(pi)=0 is y=(8sin(t))/(t^3)