<sol>:The equation is linear,thus,the characteristic equation is below:$$\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \frac { dx }{ 1 } =\frac { dy }{ 3 } =\frac { du }{ xy } \\ \quad \quad \quad \quad \quad Solve\quad theO.D.E:\\ \quad \quad \quad \quad \quad \quad \quad \quad y=3x+C\\ \quad \quad \quad \quad \quad \quad \quad \quad (xy)dx=du\quad →\quad dx(x(3x+C))=du\\ \quad \quad \quad \quad \quad hence,u=x^{ 3 }+\frac { c{ x }^{ 2 } }{ 2 } +K,\quad whereC,\quad K\quad are\quad constants.\\ \quad \quad \quad \quad \quad solve\quad IVP:\\ \quad \quad \quad \quad \quad \quad \quad \quad K=0\\ \quad \quad \quad \quad \quad \quad \quad \quad C=-3x+y\\ \quad \quad \quad \quad \quad \quad \quad \quad u=x^{ 3 }+x^{ 2 }\frac { (-3x+y) }{ 2 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =-\frac { { x }^{ 3 } }{ 2 } +\frac { y{ x }^{ 2 } }{ 2 } \\ \quad \quad \quad \quad check:\\ \quad \quad \quad \quad \quad \quad \quad \quad u_{ x }=-\frac { 3{ x }^{ 2 } }{ 2 } +xy\\ \quad \quad \quad \quad \quad \quad \quad \quad u_{ y }=\frac { { x }^{ 2 } }{ 2 } \\ \quad \quad \quad \quad \quad \quad \quad \quad u_{ x }+3u_{ y }=xy\\ \\ $$
I just try whether what I typed could be shown
Yes--but you need to surround it by double dollars (I did it for you) . Still this is extremely bad code with all these \quad and many {} are completely unnecessary; also long pieces of text (like the whole line) should not be in formulas and shorter ones should be tagged as \text{…}. And better to break the source into lines (in the logical places, where \\ are)--V.I.
