What is the solution for y^{''}-2y^'+2y=e^{-t},y(0)=0,y^'(0)=7 ?

The solution for y^{''}-2y^'+2y=e^{-t},y(0)=0,y^'(0)=7 is y=e^t(-1/5 cos(t)+37/5 sin(t))+1/5 e^{-t}

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y^{''}-2y^'+2y=e^{-t},y(0)=0,y^'(0)=7

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{- t}, y (0) = 0, y^{^{'}} (0) = 7

y = e^{t} (- \frac{1}{5} cos (t) + \frac{37}{5} sin (t)) + \frac{1}{5} e^{- t}

Solution steps

y^{''} (t) - 2 y^{'} (t) + 2 y = e^{- t}

Solve linear ODE:

y = e^{t} (- \frac{1}{5} cos (t) + \frac{37}{5} sin (t)) + \frac{1}{5} e^{- t}

y = e^{t} (- \frac{1}{5} cos (t) + \frac{37}{5} sin (t)) + \frac{1}{5} e^{- t}

\frac{d ^{2} y}{d x ^{2}} + y = 4 x

y^{^{''}} + 5 y^{^{'}} + 6 y = 1 + e^{- t}

y^{^{''}} + 2 y^{^{'}} - 8 = 0

0.5 q^{^{''}} + 20 q^{^{'}} + \frac{1}{0.0025} q = 100

x^{^{''}} + 16 x = 14 cos (3 t)

What is the solution for y^{''}-2y^'+2y=e^{-t},y(0)=0,y^'(0)=7 ?
The solution for y^{''}-2y^'+2y=e^{-t},y(0)=0,y^'(0)=7 is y=e^t(-1/5 cos(t)+37/5 sin(t))+1/5 e^{-t}