What is the solution for y^{''}-2y^'+y=0,y(0)=2,y^'(0)=6 ?

The solution for y^{''}-2y^'+y=0,y(0)=2,y^'(0)=6 is y=2e^t+4te^t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'+y=0,y(0)=2,y^'(0)=6

y^{^{''}} - 2 y^{^{'}} + y = 0, y (0) = 2, y^{^{'}} (0) = 6

y = 2 e^{t} + 4 t e^{t}

Solution steps

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

Solve linear ODE:

y = 2 e^{t} + 4 t e^{t}

y = 2 e^{t} + 4 t e^{t}

y = x^{2}, y = x^{4}

\int \frac{x + 4}{x ^{2} + 2 x + 5} d x

\frac{d}{d x} (\frac{- x ^{2} - 7}{( x ^{2} - 7 ) ^{2}})

x \to 0 + lim (x^{21 x})

\frac{\partial}{\partial y} (\frac{3 y ^{2} - t ^{2}}{y ^{5}})

What is the solution for y^{''}-2y^'+y=0,y(0)=2,y^'(0)=6 ?
The solution for y^{''}-2y^'+y=0,y(0)=2,y^'(0)=6 is y=2e^t+4te^t