# 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.

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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