What is the solution for y^{''}-16y=32+45e^x,y(0)=0,y^'(0)=1 ?

The solution for y^{''}-16y=32+45e^x,y(0)=0,y^'(0)=1 is y=3e^{4x}+2e^{-4x}-2-3e^x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-16y=32+45e^x,y(0)=0,y^'(0)=1

y^{^{''}} - 16 y = 32 + 45 e^{x}, y (0) = 0, y^{^{'}} (0) = 1

y = 3 e^{4 x} + 2 e^{- 4 x} - 2 - 3 e^{x}

Solution steps

y^{''} (x) - 16 y = 32 + 45 e^{x}

Solve linear ODE:

y = 3 e^{4 x} + 2 e^{- 4 x} - 2 - 3 e^{x}

y = 3 e^{4 x} + 2 e^{- 4 x} - 2 - 3 e^{x}

y^{^{''}} + 25 y = 2 x cos (x)

y^{^{''}} + 4 y = sec^{3} (2 t)

12 \frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 3 y = e^{x}

y^{^{''}} + 4 y = 16 t, y (0) = 3, y^{^{'}} (0) = - 6

y^{^{''}} - 4 y = 2

What is the solution for y^{''}-16y=32+45e^x,y(0)=0,y^'(0)=1 ?
The solution for y^{''}-16y=32+45e^x,y(0)=0,y^'(0)=1 is y=3e^{4x}+2e^{-4x}-2-3e^x