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\[\frac{ dy }{ dx }=x^2e^{4x}\]
solve the differential equation
i tried using by parts

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i think that just means find the anti derivative of \(x^2e^{4x}\)and my guess is use parts twice
ya i used
but not solved let me show
i will use 7 technique
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i think you have it backwards you want to drop the power on the \(x^2\) term
and now i should choose x^3 and e^4x
no the gimmick is to reduce the power, not raise it. raising the power makes it worse
\(u=x^2, dv = e^{4x}dx, du = 2x, v =\frac{1}{4}e^{4x}\)
okay i can do it this way so something wrong in my technique right
yes. as you can see your method just gave higher and higher powers of \(x\) you want to make them lower and lower
so instead of integrating id i take derivative
but if i follow the LIPET rule it says u= log right
i forget exactly how that box method works, but it is \[\int u dv =uv-\int v du\]
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
okay got u
thx
yw

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