What is the solution for y^{''}-2y^'+5y=1+t,y(0)=0,y^'(0)=1 ?

The solution for y^{''}-2y^'+5y=1+t,y(0)=0,y^'(0)=1 is y=e^t(-7/25 cos(2t)+27/50 sin(2t))+t/5+7/25

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y^{''}-2y^'+5y=1+t,y(0)=0,y^'(0)=1

y^{^{''}} - 2 y^{^{'}} + 5 y = 1 + t, y (0) = 0, y^{^{'}} (0) = 1

y = e^{t} (- \frac{7}{25} cos (2 t) + \frac{27}{50} sin (2 t)) + \frac{t}{5} + \frac{7}{25}

Solution steps

y^{''} (t) - 2 y^{'} (t) + 5 y = 1 + t

Solve linear ODE:

y = e^{t} (- \frac{7}{25} cos (2 t) + \frac{27}{50} sin (2 t)) + \frac{t}{5} + \frac{7}{25}

y = e^{t} (- \frac{7}{25} cos (2 t) + \frac{27}{50} sin (2 t)) + \frac{t}{5} + \frac{7}{25}

y^{^{''}} + y = cos (x) + 4

y^{^{''}} - 2 y^{^{'}} + 5 = 0

y^{^{''}} - 8 y^{^{'}} + 20 y = 200 x^{2} - 39

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = e^{4 x}

y^{^{''}} - 2 y^{^{'}} - 3 y = 4

What is the solution for y^{''}-2y^'+5y=1+t,y(0)=0,y^'(0)=1 ?
The solution for y^{''}-2y^'+5y=1+t,y(0)=0,y^'(0)=1 is y=e^t(-7/25 cos(2t)+27/50 sin(2t))+t/5+7/25