What is the solution for y^{''}-3y^'+2y=1+2e^{3t} ?

The solution for y^{''}-3y^'+2y=1+2e^{3t} is y=c_{1}e^{2t}+c_{2}e^t+1/2+e^{3t}

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y^{''}-3y^'+2y=1+2e^{3t}

y^{^{''}} - 3 y^{^{'}} + 2 y = 1 + 2 e^{3 t}

y = c_{1} e^{2 t} + c_{2} e^{t} + \frac{1}{2} + e^{3 t}

Solution steps

y^{''} (t) - 3 y^{'} (t) + 2 y = 1 + 2 e^{3 t}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{t} + \frac{1}{2} + e^{3 t}

y = c_{1} e^{2 t} + c_{2} e^{t} + \frac{1}{2} + e^{3 t}

t^{2} sin (\frac{1}{2} t)

n = 0 \sum \infty n^{n} x^{n}

x \to 0 lim (\frac{3 - 3 cos ( x )}{sin ^{2} ( x )})

y = (x^{2} - 9 x + 8) 25 - x^{2}

f (x) = \frac{11}{( 1 - x ) ^{2}}

What is the solution for y^{''}-3y^'+2y=1+2e^{3t} ?
The solution for y^{''}-3y^'+2y=1+2e^{3t} is y=c_{1}e^{2t}+c_{2}e^t+1/2+e^{3t}