What is the solution for y^{''}-4y^'+3y=0,y(0)=2,y^'(0)=3 ?

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

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y^{''}-4y^'+3y=0,y(0)=2,y^'(0)=3

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = 2, y^{^{'}} (0) = 3

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

Solution steps

y^{''} (t) - 4 y^{'} (t) + 3 y = 0

Solve linear ODE:

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

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

\int_{0}^{2} 5 d x

\frac{\partial}{\partial x} (x y^{9} - x^{2} y)

x \to 6 lim (x^{2} + 13)

\frac{d y}{d x} = 2 x y^{4}

y = ln (\frac{9 + e ^{x}}{9 - e ^{x}})

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