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

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:$y_2^{\prime },y_2^{″}$
$y_2^{\prime }=vy{"}_1+v^{\prime }{y}_1$
$=-v{e}^{-x}+v^{\prime }{e}^{-x}$
$=e^{-x}(-v+v^{\prime })$
$y_2^{″}=-e^{-x}(-v+v^{\prime })+e^{-x}(-v^{\prime }+v^{″})$
$=e^{-x}(v-v^{\prime }-v^{\prime }+v^{″})$
Substitute $y_2^{\prime },y_2^{″}$ in given differential equation
$e^{-x}(v-2v^{\prime }+v^{″})+e^{-x}(-v+v^{\prime })=0$
$e^{-x}(-v^{\prime }+v^{″})=0$
$-v^{\prime }+v^{″}=0$
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