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## Homework Statement

solve the following initial condition problem

[tex]x (d/dx) y(x) + 4xy(x) = -8-y(x) [/tex]

[tex]y(4)=-6 [/tex]

## Homework Equations

## The Attempt at a Solution

first i rearranged

[tex]xy'+y+4xy=-8 [/tex]

[tex]xy'+y(1+4x)=-8 [/tex]

[tex]y'+y(1/x+4)=-8/x[/tex]

integrating factor:[tex]e^\int(1/x+4)[/tex]

[tex]e^(lnx+4x)[/tex]

[tex]x+e^4x[/tex]

multiplying integrating factor:

[tex]d/dx(x+e^4x) y'+(1/x+4)(x+e^4x)y=-8/x(x+e^4x)[/tex]

[tex]d/dx(x+e^4x])y=-8/x(x+e^4x)[/tex]

[tex]\int d/dx(x+e^4x])y=\int-8/x(x+e^4x)[/tex]

[tex](x+e^4x)y=\int(-8-8e^4x/x)[/tex]

just wondering if im going alright so far... and if i am how do you integrate

[tex] \int (8e^4x/x) [/tex]

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