What is the solution for y^{''}-5y^'+6y=e^t,y(0)=1,y^'(0)=1 ?

The solution for y^{''}-5y^'+6y=e^t,y(0)=1,y^'(0)=1 is y=(-1+1/2)e^{3t}+e^{2t}+1/2 e^t

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y^{''}-5y^'+6y=e^t,y(0)=1,y^'(0)=1

y^{^{''}} - 5 y^{^{'}} + 6 y = e^{t}, y (0) = 1, y^{^{'}} (0) = 1

y = (- 1 + \frac{1}{2}) e^{3 t} + e^{2 t} + \frac{1}{2} e^{t}

Solution steps

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

Solve linear ODE:

y = (- 1 + \frac{1}{2}) e^{3 t} + e^{2 t} + \frac{1}{2} e^{t}

y = (- 1 + \frac{1}{2}) e^{3 t} + e^{2 t} + \frac{1}{2} e^{t}

What is the solution for y^{''}-5y^'+6y=e^t,y(0)=1,y^'(0)=1 ?
The solution for y^{''}-5y^'+6y=e^t,y(0)=1,y^'(0)=1 is y=(-1+1/2)e^{3t}+e^{2t}+1/2 e^t