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

The solution for y^{''}-6y^'+9y=0,y^'(0)=1 is y=(e^{3t}(1-c_{2}))/3+c_{2}e^{3t}t

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

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

y = \frac{e ^{3 t} ( 1 - c _{2} )}{3} + c_{2} e^{3 t} t

Solution steps

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

Solve linear ODE:

y = \frac{e ^{3 t} ( 1 - c _{2} )}{3} + c_{2} e^{3 t} t

y = \frac{e ^{3 t} ( 1 - c _{2} )}{3} + c_{2} e^{3 t} t

What is the solution for y^{''}-6y^'+9y=0,y^'(0)=1 ?
The solution for y^{''}-6y^'+9y=0,y^'(0)=1 is y=(e^{3t}(1-c_{2}))/3+c_{2}e^{3t}t