What is the solution for y^{''}+y^'-3y=1 ?

The solution for y^{''}+y^'-3y=1 is y=c_{1}e^{((-1+sqrt(13))t)/2}+c_{2}e^{((-1-sqrt(13))t)/2}-1/3

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y^{''}+y^'-3y=1

y^{^{''}} + y^{^{'}} - 3 y = 1

y = c_{1} e^{\frac{( - 1 + 13 ) t}{2}} + c_{2} e^{\frac{( - 1 - 13 ) t}{2}} - \frac{1}{3}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{( - 1 + 13 ) t}{2}} + c_{2} e^{\frac{( - 1 - 13 ) t}{2}} - \frac{1}{3}

y = c_{1} e^{\frac{( - 1 + 13 ) t}{2}} + c_{2} e^{\frac{( - 1 - 13 ) t}{2}} - \frac{1}{3}

y^{^{''}} + 3 y^{^{'}} + 2 y = 2 cos^{2} (x)

y^{^{''}} - 24 y = 2 x cos (5 x)

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = 2 x^{2} + e^{x} + 2 x e^{x} + 4 e^{3 x}

y^{^{''}} + y = sec (3 x)

y^{^{''}} + 3 y^{^{'}} + 2 y = 7 e^{8 x}, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}+y^'-3y=1 ?
The solution for y^{''}+y^'-3y=1 is y=c_{1}e^{((-1+sqrt(13))t)/2}+c_{2}e^{((-1-sqrt(13))t)/2}-1/3