# On solution of the differetial equation y''+y'=0 is y=e^{-x}. Use Reduction of Order to find a second linearly independent solution.

On solution of the differetial equation ${y}^{″}+{y}^{\prime }=0$ is $y={e}^{-x}$. Use Reduction of Order to find a second linearly independent solution.
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In such problems ie second order linear differential equations when we are given one solution, ${y}_{1}$ we assume the second solution to be of the form, ${y}_{2}=v{y}_{1}$ and substitute ${y}_{2}$ in the given ode and reduce order of the differential equation by using the fact that ${y}_{1}$ is a solution.
Compute:${y}_{2}^{\prime },{y}_{2}^{″}$
${y}_{2}^{\prime }=vy{"}_{1}+{v}^{\prime }{y}_{1}$
$=-v{e}^{-x}+{v}^{\prime }{e}^{-x}$
$={e}^{-x}\left(-v+{v}^{\prime }\right)$
${y}_{2}^{″}=-{e}^{-x}\left(-v+{v}^{\prime }\right)+{e}^{-x}\left(-{v}^{\prime }+{v}^{″}\right)$
$={e}^{-x}\left(v-{v}^{\prime }-{v}^{\prime }+{v}^{″}\right)$
Substitute ${y}_{2}^{\prime },{y}_{2}^{″}$ in given differential equation
${e}^{-x}\left(v-2{v}^{\prime }+{v}^{″}\right)+{e}^{-x}\left(-v+{v}^{\prime }\right)=0$
${e}^{-x}\left(-{v}^{\prime }+{v}^{″}\right)=0$
$-{v}^{\prime }+{v}^{″}=0$
Let $u={v}^{\prime }$ we get a first order differential equation
$u+{u}^{\prime }=0$
${u}^{\prime }={u}^{\prime }$
$u={e}^{x}$
Substitute $u={v}^{\prime }$ in above equation and solve for v
${v}^{\prime }={e}^{x}$
Integrating we get
$v={e}^{x}$
Get second linearly independent solution by substituting v in expression for ${y}_{2}$
${y}_{2}=v{e}^{-x}$
$=c{e}^{x}{e}^{-x}$
$=c$
Hence second linearly independent solution is c