What is the solution for y^{''}-4y^'+3y=0,y(0)=-4,y^'(0)=5 ?

The solution for y^{''}-4y^'+3y=0,y(0)=-4,y^'(0)=5 is y=(-4+17/2)e^{3t}-17/2 e^t

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y^{''}-4y^'+3y=0,y(0)=-4,y^'(0)=5

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = - 4, y^{^{'}} (0) = 5

y = (- 4 + \frac{17}{2}) e^{3 t} - \frac{17}{2} e^{t}

Solution steps

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

Solve linear ODE:

y = (- 4 + \frac{17}{2}) e^{3 t} - \frac{17}{2} e^{t}

y = (- 4 + \frac{17}{2}) e^{3 t} - \frac{17}{2} e^{t}

\frac{\partial}{\partial x} (33 - 3 x^{2} - 3 y^{2})

f (x) = \frac{x}{x ^{2} - 1}, at x = 2

(5 x^{6} + 2 x^{3})^{4}

f (x) = (5 + x) (12 - x) - 2.2 x

\frac{d}{d x} (sec^{2} (π x))

What is the solution for y^{''}-4y^'+3y=0,y(0)=-4,y^'(0)=5 ?
The solution for y^{''}-4y^'+3y=0,y(0)=-4,y^'(0)=5 is y=(-4+17/2)e^{3t}-17/2 e^t