What is the solution for y^{''}-0.2y^'+0.01y=0,y(0)=11,y^'(0)=b ?

The solution for y^{''}-0.2y^'+0.01y=0,y(0)=11,y^'(0)=b is y=11e^{t/(10)}+(b-11/10)e^{t/(10)}t

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y^{''}-0.2y^'+0.01y=0,y(0)=11,y^'(0)=b

y^{^{''}} - 0.2 y^{^{'}} + 0.01 y = 0, y (0) = 11, y^{^{'}} (0) = b

y = 11 e^{\frac{t}{10}} + (b - \frac{11}{10}) e^{\frac{t}{10}} t

Solution steps

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

Solve linear ODE:

y = 11 e^{\frac{t}{10}} + (b - \frac{11}{10}) e^{\frac{t}{10}} t

y = 11 e^{\frac{t}{10}} + (b - \frac{11}{10}) e^{\frac{t}{10}} t

\frac{d}{d x} (e^{x z})

\frac{d y}{d x} + 3 y = e^{x}

\frac{d}{d x} (\frac{5 - x e ^{x}}{x + e ^{x}})

1 + e^{x}

f (x) = 5 x^{2}, at x = 2

What is the solution for y^{''}-0.2y^'+0.01y=0,y(0)=11,y^'(0)=b ?
The solution for y^{''}-0.2y^'+0.01y=0,y(0)=11,y^'(0)=b is y=11e^{t/(10)}+(b-11/10)e^{t/(10)}t