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

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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

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