Question

Solve the differential equation by separation of variables.

(x^3 + 2)y = x(y^4 + 3)y’ ; at x = 1 and y = 1

Answer 1

Let us solve the differential equation by separation of variables:

$(x^3 + 2)y = x(y^4 + 3)y'$; at $x = 1$ and $y = 1$. It follows that $(x^3 + 2)y = x(y^4 + 3)\frac{dy}{dx},$ and hence $\frac{x^3 + 2}{x}dx = \frac{y^4 + 3}{y}dy.$ It follows that $\int\frac{x^3 + 2}{x}dx = \int\frac{y^4 + 3}{y}dy,$ and hence $\int(x^2+\frac{2}{x})dx = \int(y^3+\frac{3}{y})dy.$ Therefore, the general solution is $\frac{x^3}{3}+2\ln|x|=\frac{y^4}4+3\ln|y|+C.$ Since $y=1$ for $x=1$, we get that $\frac{1}{3}=\frac{1}4+C,$ and thus $C=\frac{1}{3}-\frac{1}4=\frac{1}{12}.$ We concludwe that the solution is $\frac{x^3}{3}+2\ln|x|=\frac{y^4}4+3\ln|y|+\frac{1}{12}.$