second order, constant coeff. homogeneous, one root repeated \[ y^{\prime \prime }-2y^{\prime }+1=0 \] Let \(y=Ae^{rx}\) and plug into the above and simplify, we obtain the charaterstic equation \begin {align*} r^{2}-2r+1 & =0\\ \left ( r-1\right ) ^{2} & =0\\ r & =1 \end {align*}
Repated root. Hence the two L.I. basis solutions are\begin {align*} y_{1} & =e^{x}\\ y_{2} & =xe^{x} \end {align*}
And the homogeneous solution is\begin {align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}e^{x}+c_{2}e^{x} \end {align*}
second order, constant coeff. homogeneous, two real distinct roots \[ y^{\prime \prime }+y^{\prime }-2y=0 \] Let \(y=Ae^{rx}\) and plug into the above and simplify, we obtain the charaterstic equation\begin {align*} r^{2}+r-2 & =0\\ \left ( r-1\right ) \left ( r+2\right ) & =0\\ r_{1} & =1\\ r_{2} & =-2 \end {align*}
Hence the L.I. basis solutions are\begin {align*} y_{1} & =e^{x}\\ y_{2} & =e^{-2x} \end {align*}
And the homogeneous solution is\begin {align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}e^{x}+c_{2}e^{-2x} \end {align*}
second order, constant coeff. homogeneous, two complex conjugate roots \[ y^{\prime \prime }-6y^{\prime }+13y=0 \] Let \(y=Ae^{rx}\) and plug into the above and simplify, we obtain the charaterstic equation\[ r^{2}-6r+13=0 \] Whose roots are \begin {align*} r_{1} & =3+2i\\ r_{2} & =3-2i \end {align*}
Hence the L.I. basis solutions are\begin {align*} y_{1} & =e^{\left ( 3+2i\right ) x}\\ y_{2} & =e^{\left ( 3-2i\right ) x} \end {align*}
the homogeneous solution is\begin {align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}e^{\left ( 3+2i\right ) x}+c_{2}e^{\left ( 3-2i\right ) x} \end {align*}
This can be converted to real basis using Euler relation which results in\begin {align*} y_{h} & =C_{1}e^{3x}\cos 2x+C_{2}e^{3}\sin 2x\\ & =e^{3x}\left ( C_{1}\cos 2x+C_{2}\sin 2x\right ) \end {align*}