What is the solution for y^{''}-5y^'=0,y^'(0)=2 ?

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

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y^{''}-5y^'=0,y^'(0)=2

y^{^{''}} - 5 y^{^{'}} = 0, y^{^{'}} (0) = 2

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

Solution steps

y^{''} (t) - 5 y^{'} (t) = 0

Solve linear ODE:

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

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

y^{^{''}} + (a + 2) y^{^{'}} + 4 y = 0

y^{^{''}} - 5 y^{^{'}} + 6.25 y = 0

3 y^{^{''}} - 5 y = 0

u^{^{''}} - 16 a^{2} H (t) = 0

y^{^{''}} + 14 y^{^{'}} + 49 y = 0, y (- 4) = - 1, y (- 4) = 5

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