What is the solution for -y^{''}+y=0,y(0)=1,y(1)=e ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-y^{''}+y=0,y(0)=1,y(1)=e

- y^{^{''}} + y = 0, y (0) = 1, y (1) = e

y = e^{t}

Solution steps

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

Solve linear ODE:

y = e^{t}

y = e^{t}

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 6

\frac{d ^{2} y}{d x ^{2}} = 9 \cdot y

y^{^{''}} - y = 0, y (0) = 4, y^{^{'}} (0) = 4

y^{^{''}} - y^{^{'}} - 120 y = 0

c \cdot y^{^{''}} + y^{^{'}} = 0

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