What is the solution for y^{''}-0.2y^'+0.01y=0,y(0)=12,y^'(0)=b ?

The solution for y^{''}-0.2y^'+0.01y=0,y(0)=12,y^'(0)=b is y=12e^{t/(10)}+(b-6/5)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)=12,y^'(0)=b

y^{^{''}} - 0.2 y^{^{'}} + 0.01 y = 0, y (0) = 12, y^{^{'}} (0) = b

y = 12 e^{\frac{t}{10}} + (b - \frac{6}{5}) e^{\frac{t}{10}} t

Solution steps

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

Solve linear ODE:

y = 12 e^{\frac{t}{10}} + (b - \frac{6}{5}) e^{\frac{t}{10}} t

y = 12 e^{\frac{t}{10}} + (b - \frac{6}{5}) e^{\frac{t}{10}} t

\frac{d}{d x} ((\frac{x ^{2}}{8} + x - \frac{1}{x})^{4})

\frac{9}{t ^{2}} - \frac{3}{t}

\int sin^{5} (5 x) cos^{3} (5 x) d x

\frac{\partial}{\partial x} (\frac{- 6 x - y}{y ^{2} + 2 x ^{2}})

\frac{d}{d x} (- \frac{4 x}{( - x + 1 ) ^{2} ( x + 1 ) ^{2}})

What is the solution for y^{''}-0.2y^'+0.01y=0,y(0)=12,y^'(0)=b ?
The solution for y^{''}-0.2y^'+0.01y=0,y(0)=12,y^'(0)=b is y=12e^{t/(10)}+(b-6/5)e^{t/(10)}t