What is the solution for y^{''}-4y^'+3y=0,y(0)=1,y^'(0)=0 ?

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

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y^{''}-4y^'+3y=0,y(0)=1,y^'(0)=0

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = 1, y^{^{'}} (0) = 0

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

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

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

\frac{\partial}{\partial x} (sin (π (6 x - 6 y)))

\frac{d}{d t} (t + 3)

\frac{d}{d x} (\frac{sin ( 2 x )}{cos ( 3 x )})

x \to 0 lim (e^{(- 3 x)})

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