# Find the general solution of the differential equation y'''-y''+y'-y=0

Find the general solution of the differential equation
$y-y+y-y=0$
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broliY

We have the differential equation
$y{}^{‴}-y{}^{″}+{y}^{\prime }-y=0$
We immediately see that it is second-order homogeneous linear DE with constant coefficients. The usual technique for this type of D.E is to first suppose that there is a solution of the form $y={e}^{mx}$.
Differentiate this supposition with respect to x, then we have
${y}^{\prime }=m{e}^{mx}$
and,
$y{}^{″}={m}^{2}{e}^{mx}$
and $y{}^{‴}={m}^{3}{e}^{mx}$
Substitute into the original differential equation then we obtain
${m}^{3}{e}^{mx}-{m}^{2}{e}^{mx}+m{e}^{mx}-{e}^{mx}=0$
${e}^{mx}\left({m}^{3}-{m}^{2}+m-1\right)=0$
Since ${e}^{mx}$ can not be equal 0, then we have
${m}^{3}-{m}^{2}+m-1=0$
Then, $\left(m-1\right)\left({m}^{2}+1\right)=0$
Then the roots are
${m}_{1}=1$
${m}_{2,3}=±i$
${m}_{2}=+i,$, ${m}_{3}=-i$
which are one real, two conjurate complex and distinct roots.
Then using theorem and theorem, the general solution of the differential equation $y{}^{″}+6{y}^{\prime }+13y=0$ is
$y={k}_{1}{e}^{x}+{c}_{1}{e}^{\left(i\right)x}+{c}_{2}{e}^{\left(i\right)x}$
$y={k}_{1}{e}^{x}+{c}_{1}{e}^{xi}+{c}_{2}{e}^{-\xi }$
where ${k}_{1},{c}_{1}$ and ${c}_{2}$ are an arbitrary constants.
Which simply is
$y={k}_{1}{e}^{x}+{k}_{2}\mathrm{cos}\left(x\right)+{k}_{3}\mathrm{sin}\left(x\right)$
where
${e}^{\xi }=\mathrm{cos}\left(x\right)+\mathrm{sin}\left(x\right)$
and ${e}^{-\xi }=\mathrm{cos}\left(x\right)-\mathrm{sin}\left(x\right)$
and ${k}_{2}=\left({c}_{1}+{c}_{2}\right),{k}_{3}=\left({c}_{1}-{c}_{2}\right)$ are a new arbitrary constants