What is the solution for y^{''}-6y^'+9y=2,y^'(0)=17 ?

The solution for y^{''}-6y^'+9y=2,y^'(0)=17 is y=(e^{3t}(17-c_{2}))/3+c_{2}e^{3t}t+2/9

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y^{''}-6y^'+9y=2,y^'(0)=17

y^{^{''}} - 6 y^{^{'}} + 9 y = 2, y^{^{'}} (0) = 17

y = \frac{e ^{3 t} ( 17 - c _{2} )}{3} + c_{2} e^{3 t} t + \frac{2}{9}

Solution steps

y^{''} (t) - 6 y^{'} (t) + 9 y = 2

Solve linear ODE:

y = \frac{e ^{3 t} ( 17 - c _{2} )}{3} + c_{2} e^{3 t} t + \frac{2}{9}

y = \frac{e ^{3 t} ( 17 - c _{2} )}{3} + c_{2} e^{3 t} t + \frac{2}{9}

(d^{2} - 1) y = x sin^{3} (x) + cos (x)

(d^{4} + 8 d^{2} + 16) \cdot y = - 2

(d^{2} + 3 d + 2) y = e^{2} x + sin (x)

y^{^{''}} - 4 y^{^{'}} + 5 y = x^{2} e^{3} x

(d^{3} - d) \cdot y = x^{3}

What is the solution for y^{''}-6y^'+9y=2,y^'(0)=17 ?
The solution for y^{''}-6y^'+9y=2,y^'(0)=17 is y=(e^{3t}(17-c_{2}))/3+c_{2}e^{3t}t+2/9