What is the solution for 25y^{''}-30y^'+9y=0,y^1(0)=e^3 x/5 ?

The solution for 25y^{''}-30y^'+9y=0,y^1(0)=e^3 x/5 is y=c_{1}e^{(3x)/5}+c_{2}e^{(3x)/5}x

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25y^{''}-30y^'+9y=0,y^1(0)=e^3 x/5

25 y^{^{''}} - 30 y^{^{'}} + 9 y = 0, y^{1} (0) = e^{3} \frac{x}{5}

y = c_{1} e^{\frac{3 x}{5}} + c_{2} e^{\frac{3 x}{5}} x

Solution steps

25 y^{''} (x) - 30 y^{'} (x) + 9 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{3 x}{5}} + c_{2} e^{\frac{3 x}{5}} x

y = c_{1} e^{\frac{3 x}{5}} + c_{2} e^{\frac{3 x}{5}} x

4 y^{^{''}} - 4 y^{^{'}} + y = 0, y (0) = 1, y^{^{'}} (0) = - 1.5

y^{^{''}} + 5 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = - 3

(d^{4} - 16) y = 0

y^{^{''}} - \frac{2}{3} y^{^{'}} + \frac{1}{9} y = 0

2 y^{^{''}} + 2 y^{^{'}} + 8 y = 0, y (0) = 5, y^{^{'}} (0) = 0

What is the solution for 25y^{''}-30y^'+9y=0,y^1(0)=e^3 x/5 ?
The solution for 25y^{''}-30y^'+9y=0,y^1(0)=e^3 x/5 is y=c_{1}e^{(3x)/5}+c_{2}e^{(3x)/5}x