What is the solution for y^{''}+4y^'=-32(x+2),y(0)=0,y^'(0)=-10 ?

The solution for y^{''}+4y^'=-32(x+2),y(0)=0,y^'(0)=-10 is y=1-e^{-4x}-4x^2-14x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'=-32(x+2),y(0)=0,y^'(0)=-10

y^{^{''}} + 4 y^{^{'}} = - 32 (x + 2), y (0) = 0, y^{^{'}} (0) = - 10

y = 1 - e^{- 4 x} - 4 x^{2} - 14 x

Solution steps

y^{''} (x) + 4 y^{'} (x) = - 32 (x + 2)

Solve linear ODE:

y = 1 - e^{- 4 x} - 4 x^{2} - 14 x

y = 1 - e^{- 4 x} - 4 x^{2} - 14 x

3 y^{^{''}} - 4 y^{^{'}} - 5 y = (x^{2} + 1) e^{- x}

y^{^{''}} + k^{2} y = k sin (k x)

2 y^{^{''}} + y^{^{'}} + 1 = 2 e^{2} x

y^{^{''}} + 2 y^{^{'}} + y = 1 + x e^{- x}

y^{^{'''}} - 2 y^{^{'}} = 2, y^{^{''}} (0) = - 1

What is the solution for y^{''}+4y^'=-32(x+2),y(0)=0,y^'(0)=-10 ?
The solution for y^{''}+4y^'=-32(x+2),y(0)=0,y^'(0)=-10 is y=1-e^{-4x}-4x^2-14x