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, \(y_2=vy_1\) 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:\(y_2',y_2''\)

\(y_2'=vy"_1+v'y_1\)

\(=-ve^{-x}+v'e^{-x}\)

\(=e^{-x}(-v+v')\)

\(y_2''=-e^{-x}(-v+v')+e^{-x}(-v'+v'')\)

\(=e^{-x}(v-v'-v'+v'')\)

Substitute \(y_2', y_2''\) in given differential equation

\(e^{-x}(v-2v'+v'')+e^{-x}(-v+v')=0\)

\(e^{-x}(-v'+v'')=0\)

\(-v'+v''=0\)

Let \(u=v'\) we get a first order differential equation

\u+u'=0\)

\(u'=u'\)

\(u=e^x\)

Substitute \(u=v'\) in above equation and solve for v

\(v'=e^x\)

Integrating we get

\(v=e^x\)

Get second linearly independent solution by substituting v in expression for \(y_2\)

\(y_2=ve^{-x}\)

\(=ce^xe^{-x}\)

\=c\)

Hence second linearly independent solution is c

Compute:\(y_2',y_2''\)

\(y_2'=vy"_1+v'y_1\)

\(=-ve^{-x}+v'e^{-x}\)

\(=e^{-x}(-v+v')\)

\(y_2''=-e^{-x}(-v+v')+e^{-x}(-v'+v'')\)

\(=e^{-x}(v-v'-v'+v'')\)

Substitute \(y_2', y_2''\) in given differential equation

\(e^{-x}(v-2v'+v'')+e^{-x}(-v+v')=0\)

\(e^{-x}(-v'+v'')=0\)

\(-v'+v''=0\)

Let \(u=v'\) we get a first order differential equation

\u+u'=0\)

\(u'=u'\)

\(u=e^x\)

Substitute \(u=v'\) in above equation and solve for v

\(v'=e^x\)

Integrating we get

\(v=e^x\)

Get second linearly independent solution by substituting v in expression for \(y_2\)

\(y_2=ve^{-x}\)

\(=ce^xe^{-x}\)

\=c\)

Hence second linearly independent solution is c