What is the solution for y^{''}-3/10 y^'+1/50 y=0 ?

The solution for y^{''}-3/10 y^'+1/50 y=0 is y=c_{1}e^{t/5}+c_{2}e^{t/(10)}

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y^{''}-3/10 y^'+1/50 y=0

y^{^{''}} - \frac{3}{10} y^{^{'}} + \frac{1}{50} y = 0

y = c_{1} e^{\frac{t}{5}} + c_{2} e^{\frac{t}{10}}

Solution steps

y^{''} (t) - \frac{3}{10} y^{'} (t) + \frac{1}{50} y = 0

Solve linear ODE:

y = c_{1} e^{\frac{t}{5}} + c_{2} e^{\frac{t}{10}}

y = c_{1} e^{\frac{t}{5}} + c_{2} e^{\frac{t}{10}}

2 y^{^{'''}} + y^{^{'}} = 0

y^{^{'''}} - 18 y^{^{''}} + 95 y^{^{'}} - 150 y = 0

\frac{d ^{2} y}{d x ^{2}} + 11 \frac{d y}{d x} + 28 y = 0

y^{^{''''}} + y^{^{'''}} - 5 y^{^{''}} - 3 y^{^{'}} + \frac{1}{2} y = 0

y^{^{''}} - 2 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 0

What is the solution for y^{''}-3/10 y^'+1/50 y=0 ?
The solution for y^{''}-3/10 y^'+1/50 y=0 is y=c_{1}e^{t/5}+c_{2}e^{t/(10)}