Question

Differential Equation problem

Answer 1

\[x \frac{ dy }{ dx }+(4x+1)y = e^{-4x}\]

Answer 2

General solution form \[\frac{ dy }{ dx }+\frac{ (4x+1) }{ x }y = \frac{ e^{-4x} }{ x }\]

Answer 3

After solving for the integrating factor\[\int\limits_{}^{} \frac{ 4x+1 }{ x } = 4x+\ln \left| x \right|\]

Answer 4

\[e^{4x+\ln \left| x \right|} = xe^{4x}\]

Answer 5

\[\frac{ d }{ dx }(xe^{4x}y) = (\frac{ e^{-4x} }{ x })(xe^{4x})\]

Answer 6

divide by x
\[\frac{ dy }{ dx }+\frac{ 4x+1 }{ x }y=\frac{ e ^{-4x} }{ x }\]
\[I.F=e ^{\int\limits \frac{ 4x+1 }{ x }dx}=e ^{\int\limits \left( 4+\frac{ 1 }{ x } \right)dx}=e ^{4x+\ln x}=e ^{4x}*e ^{\ln x }=x e ^{4x}\]
G.S.is \[y*xe ^{4x}=\int\limits\frac{ e ^{-4x} }{ x }*x e ^{4x}dx+c\]
\[xy e ^{4x}=\int\limits dx+c\]
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Answer 7

Thank you very much it is the part that i forgot that e^-4x