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ksaimouli
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\[\frac{ dy }{ dx }=x^2e^{4x}\]

ksaimouli
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solve the differential equation

ksaimouli
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i tried using by parts

anonymous
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i think that just means find the anti derivative of \(x^2e^{4x}\)and my guess is use parts twice

ksaimouli
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ya i used

ksaimouli
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but not solved let me show

ksaimouli
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i will use 7 technique

ksaimouli
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dw:1360895129811:dw

ksaimouli
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dw:1360895181802:dw

ksaimouli
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dw:1360895191642:dw

anonymous
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i think you have it backwards
you want to drop the power on the \(x^2\) term

ksaimouli
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and now i should choose x^3 and e^4x

anonymous
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no the gimmick is to reduce the power, not raise it.
raising the power makes it worse

anonymous
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\(u=x^2, dv = e^{4x}dx, du = 2x, v =\frac{1}{4}e^{4x}\)

ksaimouli
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okay i can do it this way so something wrong in my technique right

anonymous
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yes. as you can see your method just gave higher and higher powers of \(x\)
you want to make them lower and lower

ksaimouli
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so instead of integrating id i take derivative

ksaimouli
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but if i follow the LIPET rule it says u= log right

anonymous
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i forget exactly how that box method works, but it is
\[\int u dv =uv\int v du\]

anonymous
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make \(u=x^2, du = 2xdx, dv = e^{4x}dx, v =\frac{1}{4}e^{4x}\) get
\[\int x^2e^{4x}=\frac{1}{4}e^{4x}x^2\frac{1}{2}\int e^{4x}xdx\] then do it again

ksaimouli
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okay got u

ksaimouli
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thx

anonymous
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yw