What is the solution for 5y^{''}+y^'=-10x,y(0)=0,y^'(0)=-20 ?

The solution for 5y^{''}+y^'=-10x,y(0)=0,y^'(0)=-20 is y=-350+350e^{-x/5}-5x^2+50x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5y^{''}+y^'=-10x,y(0)=0,y^'(0)=-20

5 y^{^{''}} + y^{^{'}} = - 10 x, y (0) = 0, y^{^{'}} (0) = - 20

y = - 350 + 350 e^{- \frac{x}{5}} - 5 x^{2} + 50 x

Solution steps

5 y^{''} (x) + y^{'} (x) = - 10 x

Solve linear ODE:

y = - 350 + 350 e^{- \frac{x}{5}} - 5 x^{2} + 50 x

y = - 350 + 350 e^{- \frac{x}{5}} - 5 x^{2} + 50 x

2 y^{^{''}} + 3 y^{^{'}} + y = g t

y^{^{''}} - y = 2 t, y (0) = - 2 t

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

y^{^{''}} + 2 y^{^{'}} + 5 y = 3 sin (t)

y^{^{''}} - y^{^{'}} - 6 y = 62 cos (3 x)

What is the solution for 5y^{''}+y^'=-10x,y(0)=0,y^'(0)=-20 ?
The solution for 5y^{''}+y^'=-10x,y(0)=0,y^'(0)=-20 is y=-350+350e^{-x/5}-5x^2+50x