What is the solution for t^3(dy)/(dt)+3t^2y=6cos(t),y(pi)=0 ?

The solution for t^3(dy)/(dt)+3t^2y=6cos(t),y(pi)=0 is y=(6sin(t))/(t^3)

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t^3(dy)/(dt)+3t^2y=6cos(t),y(pi)=0

t^{3} \frac{d y}{d t} + 3 t^{2} y = 6 cos (t), y (π) = 0

y = \frac{6 sin ( t )}{t ^{3}}

Solution steps

t^{3} \frac{d y}{d t} + 3 t^{2} y = 6 cos (t)

Solve linear ODE:

y = \frac{6 sin ( t )}{t ^{3}}

y = \frac{6 sin ( t )}{t ^{3}}

2 \frac{d y}{d t} - y = 4 sin (3 t)

f (x) = x^{2} + 7, (1, 8)

\int 2 sin (x) - 3 sec (x) tan (x) d x

\int \frac{t ^{5}}{( t ^{2} + 4 ) ^{2}} d t

\frac{d}{d x} (ln (arccos (x)))

What is the solution for t^3(dy)/(dt)+3t^2y=6cos(t),y(pi)=0 ?
The solution for t^3(dy)/(dt)+3t^2y=6cos(t),y(pi)=0 is y=(6sin(t))/(t^3)