What is the solution for 16y^{''}-24y^'+9y=0,y^1(0)=e^{(3x)/4} ?

The solution for 16y^{''}-24y^'+9y=0,y^1(0)=e^{(3x)/4} is y=c_{1}e^{(3x)/4}+c_{2}e^{(3x)/4}x

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16y^{''}-24y^'+9y=0,y^1(0)=e^{(3x)/4}

16 y^{^{''}} - 24 y^{^{'}} + 9 y = 0, y^{1} (0) = e^{\frac{3 x}{4}}

y = c_{1} e^{\frac{3 x}{4}} + c_{2} e^{\frac{3 x}{4}} x

Solution steps

16 y^{''} (x) - 24 y^{'} (x) + 9 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{3 x}{4}} + c_{2} e^{\frac{3 x}{4}} x

y = c_{1} e^{\frac{3 x}{4}} + c_{2} e^{\frac{3 x}{4}} x

400 y^{^{''}} - 9 y = 0

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

\frac{1}{4} y^{^{''}} + y^{^{'}} + y = 0

y^{^{''}} + 10 y^{^{'}} + 23 y = 0

4 \frac{d ^{2} y}{d x ^{2}} + 12 \frac{d y}{d x} + 9 y = 0

What is the solution for 16y^{''}-24y^'+9y=0,y^1(0)=e^{(3x)/4} ?
The solution for 16y^{''}-24y^'+9y=0,y^1(0)=e^{(3x)/4} is y=c_{1}e^{(3x)/4}+c_{2}e^{(3x)/4}x