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what does this mean... obtain the particular solution satisfying the initial condition indicated \[xy^2 dx + e^xdy = 0\] when \(x \rightarrow \infty\), \(y \rightarrow 1\)
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we see that dy/dx = (-xy^2)/e^x => dy/ y^2 = -xe^-x dx just integrate both sides use this on RHS integral (uv) = u*integral(v) - integral[u' * integral(v)] will this help?
so it's by parts?
the very same i'd guess..
hmm..ok then...
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