What is the solution for y^{''}-6y^'+9y=0,y(0)=3,y^'(0)= 25/3 ?

The solution for y^{''}-6y^'+9y=0,y(0)=3,y^'(0)= 25/3 is y=3e^{3t}-(2e^{3t}t)/3

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y^{''}-6y^'+9y=0,y(0)=3,y^'(0)= 25/3

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 3, y^{^{'}} (0) = \frac{25}{3}

y = 3 e^{3 t} - \frac{2 e ^{3 t} t}{3}

Solution steps

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

Solve linear ODE:

y = 3 e^{3 t} - \frac{2 e ^{3 t} t}{3}

y = 3 e^{3 t} - \frac{2 e ^{3 t} t}{3}

\frac{d}{d x} (\frac{x ^{4}}{4} ln (x) - \frac{x ^{4}}{16})

x \to - \infty lim (x + x^{3})

(1.2) (4.4)

\int \frac{x ^{3}}{x ^{2} - 81} d x

y (θ) = \frac{sin ( θ )}{5} + \frac{c}{θ}

What is the solution for y^{''}-6y^'+9y=0,y(0)=3,y^'(0)= 25/3 ?
The solution for y^{''}-6y^'+9y=0,y(0)=3,y^'(0)= 25/3 is y=3e^{3t}-(2e^{3t}t)/3