What is the solution for y^{''}-6y+9y=0,y^1(0)=e^{3x} ?

The solution for y^{''}-6y+9y=0,y^1(0)=e^{3x} is y=c_{1}cos(sqrt(3)x)+c_{2}sin(sqrt(3)x)

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y^{''}-6y+9y=0,y^1(0)=e^{3x}

y^{^{''}} - 6 y + 9 y = 0, y^{1} (0) = e^{3 x}

y = c_{1} cos (3 x) + c_{2} sin (3 x)

Solution steps

y^{''} (x) - 6 y + 9 y = 0

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x)

y = c_{1} cos (3 x) + c_{2} sin (3 x)

y^{^{''}} + 5 y^{^{'}} + 6 y = 0, y (0) = 2, y^{^{'}} (0) = - 5

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} - 4 y^{^{'}} + 11 y = 0

\frac{y ^{^{''}}}{y ^{^{'}}} = 1

y^{^{''}} + 2 y^{^{'}} + y = 0, y (1) = 6, y^{^{'}} (0) = 6

What is the solution for y^{''}-6y+9y=0,y^1(0)=e^{3x} ?
The solution for y^{''}-6y+9y=0,y^1(0)=e^{3x} is y=c_{1}cos(sqrt(3)x)+c_{2}sin(sqrt(3)x)