What is the solution for y^{''}-0.2y^'+0.01y=0,y(0)=10,y^'(0)=b ?

The solution for y^{''}-0.2y^'+0.01y=0,y(0)=10,y^'(0)=b is y=10e^{t/(10)}+(b-1)e^{t/(10)}t

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y^{''}-0.2y^'+0.01y=0,y(0)=10,y^'(0)=b

y^{^{''}} - 0.2 y^{^{'}} + 0.01 y = 0, y (0) = 10, y^{^{'}} (0) = b

y = 10 e^{\frac{t}{10}} + (b - 1) e^{\frac{t}{10}} t

Solution steps

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

Solve linear ODE:

y = 10 e^{\frac{t}{10}} + (b - 1) e^{\frac{t}{10}} t

y = 10 e^{\frac{t}{10}} + (b - 1) e^{\frac{t}{10}} t

\int cos (30 x) \cdot cos (4 x) \cdot cos (4 x) d x

y^{^{'}} = x + y, y (0) = 10

\frac{\partial}{\partial v} (u (v, w) + v e^{w})

n = 2 \sum \infty \frac{8}{n ln ( n )}

\frac{d y}{d x} = \frac{e ^{2 x - y}}{e ^{x + y}}

What is the solution for y^{''}-0.2y^'+0.01y=0,y(0)=10,y^'(0)=b ?
The solution for y^{''}-0.2y^'+0.01y=0,y(0)=10,y^'(0)=b is y=10e^{t/(10)}+(b-1)e^{t/(10)}t