What is the solution for y^{''}-5y^'+6y=0,y(0)=1,y^'(0)= 7/4 ?

The solution for y^{''}-5y^'+6y=0,y(0)=1,y^'(0)= 7/4 is y=(1-5/4)e^{3t}+5/4 e^{2t}

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y^{''}-5y^'+6y=0,y(0)=1,y^'(0)= 7/4

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = \frac{7}{4}

y = (1 - \frac{5}{4}) e^{3 t} + \frac{5}{4} e^{2 t}

Solution steps

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

Solve linear ODE:

y = (1 - \frac{5}{4}) e^{3 t} + \frac{5}{4} e^{2 t}

y = (1 - \frac{5}{4}) e^{3 t} + \frac{5}{4} e^{2 t}

\frac{d ^{2} y}{d x ^{2}} + 5 \cdot \frac{d y}{d x} = 0

y^{^{''}} - 2 y^{^{'}} - 15 y = 0, y (0) = 1, y^{^{'}} (0) = 37

2 u^{^{''}} + \frac{3}{5} u^{^{'}} + 30 H (t) = 0

49 y^{^{''}} - 112 y^{^{'}} + 64 y = 0

- \frac{d ^{2} y}{d x ^{2}} + 400 y = 0

What is the solution for y^{''}-5y^'+6y=0,y(0)=1,y^'(0)= 7/4 ?
The solution for y^{''}-5y^'+6y=0,y(0)=1,y^'(0)= 7/4 is y=(1-5/4)e^{3t}+5/4 e^{2t}