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

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

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

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

y = (1 - \frac{4}{3}) e^{3 t} + \frac{4}{3} e^{t}

Solution steps

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

Solve linear ODE:

y = (1 - \frac{4}{3}) e^{3 t} + \frac{4}{3} e^{t}

y = (1 - \frac{4}{3}) e^{3 t} + \frac{4}{3} e^{t}

\frac{d}{d x} (x^{4} + 8 x)

f (x) = - \frac{4}{x}

x^{3}, at x = 1

y = (x^{6} - 10) x

\frac{d y}{d t} = 0.0572 (400 - y), y (0) = 32

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