What is the solution for y^{''}-5y^'+6y=0,y(1)=e^2,y^'(1)=3e^2 ?

The solution for y^{''}-5y^'+6y=0,y(1)=e^2,y^'(1)=3e^2 is y=e^{3t-1}

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y^{''}-5y^'+6y=0,y(1)=e^2,y^'(1)=3e^2

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (1) = e^{2}, y^{^{'}} (1) = 3 e^{2}

y = e^{3 t - 1}

Solution steps

y^{''} (t) - 5 y^{'} (t) + 6 y = 0

Solve linear ODE:

y = e^{3 t - 1}

y = e^{3 t - 1}

\int x^{2} cos (x^{3} + 5) d x

\int \frac{x ^{3}}{1 + x ^{8}} d x

\frac{d}{d x} (\frac{1}{4} e^{- \frac{x}{2}})

x \to π lim (sin (x - π))

x \to 0 + lim (- \frac{3}{x})

What is the solution for y^{''}-5y^'+6y=0,y(1)=e^2,y^'(1)=3e^2 ?
The solution for y^{''}-5y^'+6y=0,y(1)=e^2,y^'(1)=3e^2 is y=e^{3t-1}