What is the solution for y^{''}-0.2y^'+0.01y=0,y(0)=15,y^'(0)=b ?

The solution for y^{''}-0.2y^'+0.01y=0,y(0)=15,y^'(0)=b is y=15e^{t/(10)}+(b-3/2)e^{t/(10)}t

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y^{''}-0.2y^'+0.01y=0,y(0)=15,y^'(0)=b

y^{^{''}} - 0.2 y^{^{'}} + 0.01 y = 0, y (0) = 15, y^{^{'}} (0) = b

y = 15 e^{\frac{t}{10}} + (b - \frac{3}{2}) e^{\frac{t}{10}} t

Solution steps

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

Solve linear ODE:

y = 15 e^{\frac{t}{10}} + (b - \frac{3}{2}) e^{\frac{t}{10}} t

y = 15 e^{\frac{t}{10}} + (b - \frac{3}{2}) e^{\frac{t}{10}} t

x d y - (2 x + 1) e^{- y} d x = 0

\int \frac{1}{( 3 x - 1 )} d x

\frac{\partial}{\partial x} (2 cos (3 x))

\frac{d y}{d x} + y^{5} x + 4 y = 0

\int_{0}^{2} x^{2} e^{- x} d x

What is the solution for y^{''}-0.2y^'+0.01y=0,y(0)=15,y^'(0)=b ?
The solution for y^{''}-0.2y^'+0.01y=0,y(0)=15,y^'(0)=b is y=15e^{t/(10)}+(b-3/2)e^{t/(10)}t