What is the solution for y^{''}-y^'=3-2x,y(0)=-1,y^'(0)=-1 ?

The solution for y^{''}-y^'=3-2x,y(0)=-1,y^'(0)=-1 is y=x^2-x-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y^'=3-2x,y(0)=-1,y^'(0)=-1

y^{^{''}} - y^{^{'}} = 3 - 2 x, y (0) = - 1, y^{^{'}} (0) = - 1

y = x^{2} - x - 1

Solution steps

y^{''} (x) - y^{'} (x) = 3 - 2 x

Solve linear ODE:

y = x^{2} - x - 1

y = x^{2} - x - 1

y^{^{''}} - 25 y = e^{x}

y^{^{''}} - 14 y^{^{'}} + 45 y = 360 x^{2} - 224 x + 16

y^{^{''}} + 4 y = 4 cos (2 x) + 6 cos (x)

y^{^{''}} + 5 y^{^{'}} + 4 y = sec (x)

y^{^{''}} + 4 y = cos (t), y^{^{'}} (0) = 1, y (0) = 1

What is the solution for y^{''}-y^'=3-2x,y(0)=-1,y^'(0)=-1 ?
The solution for y^{''}-y^'=3-2x,y(0)=-1,y^'(0)=-1 is y=x^2-x-1