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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{''}} + 16.33 x = 0

y^{^{''}} - 18 y^{^{'}} + 81 y = 0, y (0) = 1, y (1) = 0

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

y^{^{''}} - 2 y^{^{'}} - 10 y = 0

0.5 y^{^{''}} + y^{^{'}} + 10 y = 0, y (0) = 0, y^{^{'}} (0) = 1

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