What is the solution for y^{''}-0.2y^'+0.01y=0,y(0)=13,y^'(0)=b ?

The solution for y^{''}-0.2y^'+0.01y=0,y(0)=13,y^'(0)=b is y=13e^{t/(10)}+(b-13/10)e^{t/(10)}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-0.2y^'+0.01y=0,y(0)=13,y^'(0)=b

y^{^{''}} - 0.2 y^{^{'}} + 0.01 y = 0, y (0) = 13, y^{^{'}} (0) = b

y = 13 e^{\frac{t}{10}} + (b - \frac{13}{10}) e^{\frac{t}{10}} t

Solution steps

y^{''} (t) - 0.2 y^{'} (t) + 0.01 y = 0

Solve linear ODE:

y = 13 e^{\frac{t}{10}} + (b - \frac{13}{10}) e^{\frac{t}{10}} t

y = 13 e^{\frac{t}{10}} + (b - \frac{13}{10}) e^{\frac{t}{10}} t

\int 18 tan^{17} (x) sec^{2} (x) d x

x \to 9 lim (\frac{3 - x}{9 x - x ^{2}})

\int \frac{1}{u ^{2} + 3} d u

\frac{d x}{d y} = \frac{y ^{3} - 3 x}{y}

What is the solution for y^{''}-0.2y^'+0.01y=0,y(0)=13,y^'(0)=b ?
The solution for y^{''}-0.2y^'+0.01y=0,y(0)=13,y^'(0)=b is y=13e^{t/(10)}+(b-13/10)e^{t/(10)}t