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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} x}{d t ^{2}} + 0.75 \frac{d x}{d t} + 0.3 x = 0

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

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = - \frac{5}{3}

x^{^{''}} + 5 x^{^{'}} + 3 x = 0, x (0) = 5, x^{^{'}} (0) = 7

y^{^{''}} - 22 y^{^{'}} + 121 y = 0, y (0) = 0, y^{^{'}} (0) = 5

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