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)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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}}

x \to 3 + lim (3 ln (x - 2))

\int cos (3 x^{2} + 5) x d x

f (x) = x^{4}, (- 2, 16)

\int 7 x (e^{6 x^{2}} - 1) d x

\frac{d}{d x} (10000 - 40 x - 0.02 x^{2})

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)}