Respuesta :
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=6
we have [tex]6=8+4c[/tex]
Thus solving for c we get c = -1/2
Thus particular solution becomes
[tex]y=x^{3}-\frac{1}{2}x^{2}[/tex]
