What is the solution for y^{''}-y^'-12y=0,y(1)=0,y^'(1)=1 ?

The solution for y^{''}-y^'-12y=0,y(1)=0,y^'(1)=1 is y=(e^{4t-4})/7-(e^3)/7 e^{-3t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y^'-12y=0,y(1)=0,y^'(1)=1

y^{^{''}} - y^{^{'}} - 12 y = 0, y (1) = 0, y^{^{'}} (1) = 1

y = \frac{e ^{4 t - 4}}{7} - \frac{e ^{3}}{7} e^{- 3 t}

Solution steps

y^{''} (t) - y^{'} (t) - 12 y = 0

Solve linear ODE:

y = \frac{e ^{4 t - 4}}{7} - \frac{e ^{3}}{7} e^{- 3 t}

y = \frac{e ^{4 t - 4}}{7} - \frac{e ^{3}}{7} e^{- 3 t}

25 y^{^{''}} + 9 y = 0

(D^{2} - 2 D + 2) y = 0

4 \frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} = - y

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

\frac{d ^{2} y}{d x ^{2}} - 8 \frac{d y}{d x} + 15 y = 0

What is the solution for y^{''}-y^'-12y=0,y(1)=0,y^'(1)=1 ?
The solution for y^{''}-y^'-12y=0,y(1)=0,y^'(1)=1 is y=(e^{4t-4})/7-(e^3)/7 e^{-3t}