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

The solution for t^2(dy)/(dt)+y=0,y(6)=-3,t>0 is y=-3e^{(6-t)/(6t)}

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t^2(dy)/(dt)+y=0,y(6)=-3,t>0

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

y = - 3 e^{\frac{6 - t}{6 t}}

Solution steps

t^{2} \frac{d y}{d t} + y = 0

Solve linear ODE:

y = - 3 e^{\frac{6 - t}{6 t}}

y = - 3 e^{\frac{6 - t}{6 t}}

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What is the solution for t^2(dy)/(dt)+y=0,y(6)=-3,t>0 ?
The solution for t^2(dy)/(dt)+y=0,y(6)=-3,t>0 is y=-3e^{(6-t)/(6t)}