What is the solution for t^2(dy)/(dt)+y=0,y(5)=-1 ?

The solution for t^2(dy)/(dt)+y=0,y(5)=-1 is y=-e^{(5-t)/(5t)}

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t^2(dy)/(dt)+y=0,y(5)=-1

t^{2} \frac{d y}{d t} + y = 0, y (5) = - 1

y = - e^{\frac{5 - t}{5 t}}

Solution steps

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

Solve linear ODE:

y = - e^{\frac{5 - t}{5 t}}

y = - e^{\frac{5 - t}{5 t}}

\int 1. 8^{(0.045 x)} - 1 d x

\int - x^{2} e^{- x} d x

(1 - x^{2}) y^{^{''}} - 2 x y^{^{'}} = 1

x \to 0 lim (1 - 1 - x)

2 x y^{^{'}} + y = 6 x

What is the solution for t^2(dy)/(dt)+y=0,y(5)=-1 ?
The solution for t^2(dy)/(dt)+y=0,y(5)=-1 is y=-e^{(5-t)/(5t)}