Solve the following initial-value problem, showing all work, including a clear general solution as well as the particular solution requested (Label the Gen. Sol. and Particular Sol.). Write both solutions in EXPLICIT FORM (solved for y). x dy/dx = x^3 + 2y subject to: y(2) = 6

Answer:General Solution is [tex]y=x^{3}+cx^{2}[/tex] and the particular solution is  [tex]y=x^{3}-\frac{1}{2}x^{2}[/tex]Step-by-step explanation:[tex]x\frac{\mathrm{dy} }{\mathrm{d} x}=x^{3}+3y\\\\Rearranging \\\\x\frac{\mathrm{dy} }{\mathrm{d} x}-3y=x^{3}\\\\\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{3y}{x}=x^{2}[/tex]This is a linear diffrential equation of type[tex]\frac{\mathrm{d} y}{\mathrm{d} x}+p(x)y=q(x)[/tex]..................(i)here [tex]p(x)=\frac{-2}{x}[/tex][tex]q(x)=x^{2}[/tex]The solution of equation i is given by[tex]y\times e^{\int p(x)dx}=\int  e^{\int p(x)dx}\times q(x)dx[/tex]we have [tex]e^{\int p(x)dx}=e^{\int \frac{-2}{x}dx}\\\\e^{\int \frac{-2}{x}dx}=e^{-2ln(x)}\\\\=e^{ln(x^{-2})}\\\\=\frac{1}{x^{2} } \\\\\because e^{ln(f(x))}=f(x)]\\\\Thus\\\\e^{\int p(x)dx}=\frac{1}{x^{2}}[/tex]Thus the solution becomes [tex]\tfrac{y}{x^{2}}=\int \frac{1}{x^{2}}\times x^{2}dx\\\\\tfrac{y}{x^{2}}=\int 1dx\\\\\tfrac{y}{x^{2}}=x+c[/tex][tex]y=x^{3}+cx^{2[/tex]This is the general solution now to find the particular solution we put value of x=2 for which y=6we have [tex]6=8+4c[/tex]Thus solving for c we get c = -1/2Thus particular solution becomes[tex]y=x^{3}-\frac{1}{2}x^{2}[/tex]