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

5 y^{^{''}} + 80 y = 0

y^{^{''}} + 7 y^{^{'}} + 10 y = 0, y (0) = 11, y^{^{'}} (0) = - 49

4000 y^{^{''}} - 9 y^{^{'}} = 0

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

x^{^{''}} + 64 x^{^{'}} = 0

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