Practicing Success
If \(c\) is an arbitrary constant, the general solution of the differential equation \((x+y^3)\frac{dy}{dx}=y\) is |
\(2x=y^3-cy^2\) \(2x=y^3+cy\) \(y=x^3+cx\) \(y=x^3-cx^2\) |
\(2x=y^3+cy\) |
\((x+y^3)\frac{dy}{dx}=y\) $⇒\frac{(x+y^3)}{y}=\frac{dx}{dy}$ so $\frac{dx}{dy}-\frac{x}{y}=y^2$ so $I.F.=e^{\int -\frac{1}{y}dy}=e^{ln|\frac{1}{y}|}=\frac{1}{y}$ so $\int\frac{1}{y}\frac{dx}{dy}-\frac{1}{y^2}xdy=\int ydy$ $\frac{x}{y}=\frac{y^2}{2}+C$ $x=\frac{y^3}{2}+\frac{2cy}{2}$ |