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

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

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

y^{^{''}} + 4 y = 0, y (0) = 3, y^{^{'}} (0) = - 4

y^{^{''}} + 4 y = 0, y (0) = 3, y^{^{'}} (0) = - 3

5 y^{^{''}} + 5 y^{^{'}} + 45 y = 0, y (0) = - 8, y^{^{'}} (0) = 0

y^{^{''}} - \frac{10}{3} y^{^{'}} + α y = 0, y^{^{'}} (0) = - \frac{2}{3}

x^{^{''}} - x^{^{'}} - \frac{3}{4} x = 0

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