2019 JPJC P2 Q4

A differential equation is given by \(2{{u}^{2}}\frac{{{\mathrm{d}}^{2}}x}{\mathrm{d}{{u}^{2}}}+4u\frac{\mathrm{d}x}{\mathrm{d}u}=15u+12\) where \(x=0\) and \(\frac{\mathrm{d}x}{\mathrm{d}u}=1\) when \(u=1\).

By differentiating \({{u}^{2}}\frac{\mathrm{d}x}{\mathrm{d}u}\) with respect to \(u\), show that the solution of the differential equation is given by \(x=au+\frac{b}{u}+c\mathrm{f}\left( u \right)+d\), where \(a,b,c\) and \(d\) are constants to be determined and \(\mathrm{f}\left( u \right)\) is a function in \(u\) to be found.[6]