Question

Solve differential equation \(u'-5u=ve^{-5v}\)

Answer 1

\(\frac{du}{dv}-5u= ve^{-5v}\)
\(\frac{dy}{dx}+Py=Q\)
\(I.F.= e^{\int Pdx}\)
\(= e^{- \int 5dv}\)
\(= e^{-5 \int dv}\)
\(= e^{-5v}\)
\(u(I.F.)= \int ve^{-5v}(I.F.)dv+c\)
\(ue^{-5v}= \int ve^{-5v} e^{-5v}dv+c\)
\(ue^{-5v}= \int ve^{-10v}dv+c\) (1)
Now find \(\int ve^{-10v}dv\) (by parts)
\(= v \int e^{-10v}dv- \int [\frac{d}{dv}(v) \int e^{-10v}dv]dv\)
\(= v \frac{e^{-10v}}{-10} - \int [\frac{e^{-10v}}{-10}]dv\)
\(= \frac{ve^{-10v}}{-10}+\frac{1}{10}\frac{e^{-10v}}{-10}\) (2)
from (1) and (2)
\(ue^{-5v}= \frac{ve^{-10v}}{-10}-\frac{1}{100} e^{-10v}+c\)
\(u=( \frac{ve^{-10v}}{-10} e^{-5v})-\frac{1}{100}\frac{e^{-10v}}{e^{-5v}}+\frac{c}{e^{-5v}}\)
\(u= \frac{ve^{-5v}}{-10}-\frac{1}{100} e^{-5v}+ce^{5v}\)