Answer: y(x) = C.x.[tex]e^{\frac{3x}{2}+x}[/tex]
Step-by-step explanation: To solve the differential equation:
1) Put similar terms together:
[tex]\frac{dy}{dx} = \frac{y}{x}+3x+1[/tex]
[tex]\frac{dy}{y} = (\frac{1}{x} +3x + 1) dx[/tex]
2) Integrate both sides
[tex]\int\limits {\frac{dy}{y} } = \int\ {\frac{1}{x} + 3x + 1 } \, dx[/tex]
ln y = ln x + [tex]\frac{3}{2}x^{2}[/tex] + x + c
y(x) = [tex]e^{lnx + \frac{3}{2}x^{2}+x}.e^{c}[/tex]
3) Knowing that c is a constant and [tex]e^{lnx} = x[/tex]:
y(x) = C.x.[tex]e^{\frac{3}{2}.x^{2} + x }[/tex]
The general solution to the equation is: y(x) = C.x.[tex]e^{\frac{3}{2}.x^{2} + x }[/tex]
