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

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

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

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

y = 2 e^{3 t} - 3 e^{t}

Solution steps

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

Solve linear ODE:

y = 2 e^{3 t} - 3 e^{t}

y = 2 e^{3 t} - 3 e^{t}

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = - 3, y^{^{'}} (0) = 1

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 2, y^{^{'}} (0) = 5

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = 4, y^{^{'}} (0) = - 1

y^{^{''}} - 25 y^{^{'}} + 144 y = 0

x^{^{''}} + 3 x^{^{'}} + x = 0

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