What is the solution for f^{''}(t)-4f^'(t)+3f(t)=0,f(0)=3,f^'(0)=5 ?

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

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f^{''}(t)-4f^'(t)+3f(t)=0,f(0)=3,f^'(0)=5

f^{^{''}} (t) - 4 f^{^{'}} (t) + 3 f (t) = 0, f (0) = 3, f^{^{'}} (0) = 5

f (t) = e^{3 t} + 2 e^{t}

Solution steps

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

Solve linear ODE:

f (t) = e^{3 t} + 2 e^{t}

f (t) = e^{3 t} + 2 e^{t}

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} + 7 y = 0

5 y^{^{''}} + 50 y^{^{'}} + 125 y = 0

\frac{d ^{3} y}{d x ^{3}} - \frac{d y}{d x} = 0

y^{^{''}} + a y = 0, y (0) = a, y^{^{'}} (0) = 0

\frac{1}{8} y^{^{''}} - 2 y = 0

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