Answer to Question #249870 in Differential Equations for Jia

Question

exact dx+(x^2y+4y)dy=0 y(4)=0

Answer 1

Let us solve the differential equation $dx+(x^2y+4y)dy=0,\ y(4)=0.$

Since $\frac{\partial (1)}{\partial y}=0$ but $\frac{\partial (x^2y+4y)}{\partial x}=2xy\ne 0,$ we conclude that this differential equation is not exact.

Let us solve it. For this let us divide both parts by $x^2+4.$ Then we have the equivalent differential equation $\frac{dx}{x^2+4}+ydy=0.$ It follows that $\int\frac{dx}{x^2+4}+\int ydy=C,$ and hence $\frac{1}2\arctan\frac{x}2+\frac{y^2}2=C.$

Taking into account that $y(4)=0,$ we conclude that $C=\frac{1}2\arctan\frac{4} 2+0=\frac{1}2\arctan 2.$

Therefore, $\arctan\frac{x}2+y^2=\arctan 2$ is the solution of the differential equation $dx+(x^2y+4y)dy=0,\ y(4)=0.$