What is the solution for y^{''}+3y^'=0,y(0)=2,y^'(0)=-3 ?

The solution for y^{''}+3y^'=0,y(0)=2,y^'(0)=-3 is y=e^{-3t}+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+3y^'=0,y(0)=2,y^'(0)=-3

y^{^{''}} + 3 y^{^{'}} = 0, y (0) = 2, y^{^{'}} (0) = - 3

y = e^{- 3 t} + 1

Solution steps

y^{''} (t) + 3 y^{'} (t) = 0

Solve linear ODE:

y = e^{- 3 t} + 1

y = e^{- 3 t} + 1

\int \frac{1}{1 - 3 y} d y

y^{^{''}} + 2 y^{^{'}} = 8 sin (2 t) + 32 cos (2 t)

f (x) = 5 x^{2} - 6

f (x) = 4 x^{2} + 2 x, (- 1, 2)

- \frac{1}{2 x ^{2}} + \frac{32}{5 x ^{5}}

What is the solution for y^{''}+3y^'=0,y(0)=2,y^'(0)=-3 ?
The solution for y^{''}+3y^'=0,y(0)=2,y^'(0)=-3 is y=e^{-3t}+1