What is the solution for y^{''}-5y^'+4y=e^{2t},y(0)=1,y^'(0)=-1 ?

The solution for y^{''}-5y^'+4y=e^{2t},y(0)=1,y^'(0)=-1 is y=(-2+3/2)e^{4t}+2e^t-1/2 e^{2t}

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y^{''}-5y^'+4y=e^{2t},y(0)=1,y^'(0)=-1

y^{^{''}} - 5 y^{^{'}} + 4 y = e^{2 t}, y (0) = 1, y^{^{'}} (0) = - 1

y = (- 2 + \frac{3}{2}) e^{4 t} + 2 e^{t} - \frac{1}{2} e^{2 t}

Solution steps

y^{''} (t) - 5 y^{'} (t) + 4 y = e^{2 t}

Solve linear ODE:

y = (- 2 + \frac{3}{2}) e^{4 t} + 2 e^{t} - \frac{1}{2} e^{2 t}

y = (- 2 + \frac{3}{2}) e^{4 t} + 2 e^{t} - \frac{1}{2} e^{2 t}

What is the solution for y^{''}-5y^'+4y=e^{2t},y(0)=1,y^'(0)=-1 ?
The solution for y^{''}-5y^'+4y=e^{2t},y(0)=1,y^'(0)=-1 is y=(-2+3/2)e^{4t}+2e^t-1/2 e^{2t}