- #1

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you are given that [itex]y(0)=4[/itex].

so, here's what i did:

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

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

i integrated both sides:

[tex]ln(y+1)=\frac{x^2}{2}-4x+C[/tex]

[tex]y=e^{\frac{x^2}{2}}e^{-4x}e^{C}-1[/tex]

plugged in for x and y:

[tex]4=e^{\frac{0^2}{x}}e^{-4*0}e^{C}-1[/tex]

[tex]5=e^C[/tex]

so:

[tex]y=5e^{\frac{x^2}{x}}e^{-4x}[/tex]

i know that's wrong, and i need help working it out. that's what i have so far though.