91影视

Let the given function be \(y = {c_1}{x^{ - 1}} + {c_2}x + {c_3}xlnx + 4{x^2}\).

Then, the first derivative of the function is,

\(\frac{{dy}}{{dx}} = - {c_1}{x^{ - 2}} + {c_2} + {c_3}(lnx + 1) + 8x\)

The second derivative of the function is,

\(\frac{{{d^2}y}}{{d{x^2}}} = 2{c_1}{x^{ - 3}} + {c_3}{x^{ - 1}} + 8\)

The third derivative of the function is,

\(\frac{{{d^3}y}}{{d{x^3}}} = - 6{c_1}{x^{ - 4}} - {c_3}{x^{ - 2}}\)

Substitute \(y\), \(y'\) and \(y''\) into the left-hand side of the differential equation.

\(\begin{aligned}{c}{x^3}\left( { - 6{c_1}{x^{ - 4}} - {c_3}{x^{ - 2}}} \right) + 2{x^2}\left( {2{c_1}{x^{ - 3}} + {c_3}{x^{ - 1}} + 8} \right) - x\left( { - {c_1}{x^{ - 2}} + {c_2} + {c_3}(lnx + 1) + 8x} \right) + \left( {{c_1}{x^{ - 1}} + {c_2}x + {c_3}xlnx + 4{x^2}} \right) &= 12{x^2}\\{c_1}\left( { - 6{x^{ - 1}} + 4{x^{ - 1}} + {x^{ - 1}} + {x^{ - 1}}} \right) + {c_2}( - x + x) + {c_3}( - x + 2x - xlnx - x + xlnx) + \left( {16{x^2} - 8{x^2} + 4{x^2}} \right) &= 12{x^2}\\12{x^2} &= 12{x^2}\end{aligned}\)

That is same as the right-hand side of the differential equation. As the solution has logarithmic term, the solution is defined when \(x > 0\), (i.e.) \(I:(0,\infty )\). Thus, the indicated function is a solution of the given differential equation.