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

The solution for y^{''}-6y^'+9y=0,y(1/3)=9,y(0)=3 is y=((27-(-9e+27))e^{3t-1})/3+(-9e+27)e^{3t-1}t

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

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

y = \frac{( 27 - ( - 9 e + 27 ) ) e ^{3 t - 1}}{3} + (- 9 e + 27) e^{3 t - 1} t

Solution steps

y^{''} (t) - 6 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = \frac{( 27 - ( - 9 e + 27 ) ) e ^{3 t - 1}}{3} + (- 9 e + 27) e^{3 t - 1} t

y = \frac{( 27 - ( - 9 e + 27 ) ) e ^{3 t - 1}}{3} + (- 9 e + 27) e^{3 t - 1} t

What is the solution for y^{''}-6y^'+9y=0,y(1/3)=9,y(0)=3 ?
The solution for y^{''}-6y^'+9y=0,y(1/3)=9,y(0)=3 is y=((27-(-9e+27))e^{3t-1})/3+(-9e+27)e^{3t-1}t