What is the solution for y^{''}-4y=8e^{2x},y(0)=1,y^'(0)=0 ?

The solution for y^{''}-4y=8e^{2x},y(0)=1,y^'(0)=0 is y=2e^{2x}x+e^{-2x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-4y=8e^{2x},y(0)=1,y^'(0)=0

y^{^{''}} - 4 y = 8 e^{2 x}, y (0) = 1, y^{^{'}} (0) = 0

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

Solution steps

y^{''} (x) - 4 y = 8 e^{2 x}

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 40 y = 25 sin (6 t)

x^{2} y^{^{''}} - 8 x^{2} y^{^{'}} + 16 x^{2} y = 3 e^{4 x}

y^{^{''}} + 3 y^{^{'}} + 2 y = 12 e^{t}, y (0) = 2, y^{^{'}} (0) = 3

y^{^{''}} + 4 y^{^{'}} + 3 y = 4 x^{2} + 5

y^{^{''}} - y^{^{'}} - 6 y = 73 cos (3 x)

What is the solution for y^{''}-4y=8e^{2x},y(0)=1,y^'(0)=0 ?
The solution for y^{''}-4y=8e^{2x},y(0)=1,y^'(0)=0 is y=2e^{2x}x+e^{-2x}