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

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

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

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

y = (3 - \frac{7}{2}) e^{3 t} + \frac{7}{2} e^{t}

Solution steps

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

Solve linear ODE:

y = (3 - \frac{7}{2}) e^{3 t} + \frac{7}{2} e^{t}

y = (3 - \frac{7}{2}) e^{3 t} + \frac{7}{2} e^{t}

x^{2} + 3 x

\frac{d}{d x} (5 x lo g_{4} (2 x - 1))

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

a

y^{^{''}} - 8 y^{^{'}} + 14 y = 0, y (0) = 0, y^{^{'}} (0) = 1

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