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

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

Answers (1)

2021-02-13
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
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