$x^4y'' + 2x^3y' - y = 16e^{\frac{3}{x}}$ Ans: $y = c_1e^{\frac{1}{x}} + c_2e^{-\frac{1}{x}} + 2e^{\frac{3}{x}}$
Added by Russell E.
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Let's assume a solution of the form $y = x^m$. Then $y' = mx^{m-1}$ and $y'' = m(m-1)x^{m-2}$. Substituting into the homogeneous equation ($x^4y'' + 2x^3y' - y = 0$), we get: $x^4(m(m-1)x^{m-2}) + 2x^3(mx^{m-1}) - x^m = 0$ $m(m-1)x^{m+2} + 2mx^{m+2} - x^m = Show more…
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