What is the solution for y^{''}-2y^'+1y=0,y(0)=5,y^'(0)=4 ?

The solution for y^{''}-2y^'+1y=0,y(0)=5,y^'(0)=4 is y=5e^t-e^tt

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'+1y=0,y(0)=5,y^'(0)=4

y^{^{''}} - 2 y^{^{'}} + 1 y = 0, y (0) = 5, y^{^{'}} (0) = 4

y = 5 e^{t} - e^{t} t

Solution steps

y^{''} (t) - 2 y^{'} (t) + 1 y = 0

Solve linear ODE:

y = 5 e^{t} - e^{t} t

y = 5 e^{t} - e^{t} t

ln (x - 9)

\frac{\partial}{\partial x} ((x + 1)^{2} (x^{2} + 2 x))

f (x) = \frac{x ^{3} - 1 x ^{2} + 2}{x ^{2}}

\int (x x + 3) d x

\frac{\partial}{\partial x} (x^{4} + y^{4} + 3 x^{2} y^{3})

What is the solution for y^{''}-2y^'+1y=0,y(0)=5,y^'(0)=4 ?
The solution for y^{''}-2y^'+1y=0,y(0)=5,y^'(0)=4 is y=5e^t-e^tt