Answer to Question #246811 in Differential Equations for weiou

Question

determine the exactness and obtain the general solutions.

1. (𝑥 + 𝑦² )𝑑𝑥 + (2𝑥𝑦 + 𝑦²)𝑑𝑦 = 0

2. (2𝑥𝑦³ + 𝑥² )𝑑𝑥 + (3𝑥²𝑦² + 𝑦)𝑑𝑦 = 0

3. (𝑒^2y + 𝑥)𝑑𝑥 + (2𝑥𝑒^2y + 1)𝑑𝑦 = 0 

Answer 1

Let us determine the exactness and obtain the general solutions.

1. Consider the equation $(𝑥 + 𝑦^2 )𝑑𝑥 + (2𝑥𝑦 + 𝑦^2)𝑑𝑦 = 0$. Since $\frac{{\partial (x+y^2)}}{\partial y}=2y=\frac{{\partial (2xy+y^2)}}{\partial x},$ we conclude that the differential equation is exact. Then $\frac{{\partial U}}{\partial x}=x+y^2, \frac{{\partial U}}{\partial y}=2xy+y^2.$ It follows that $U=\frac{x^2}2+y^2x+C(y),$ and hence $\frac{{\partial U}}{\partial y}=2yx+C'(y)=2xy+y^2.$ Therefore, $C'(y)=y^2,$ and thus $C(y)=\frac{y^3}3+C.$ We conclude that the general solution of the differential equation is $\frac{x^2}2+y^2x+\frac{y^3}3=C.$

2. Consider the equation $(2𝑥𝑦^3 + 𝑥^2 )𝑑𝑥 + (3𝑥^2𝑦^2 + 𝑦)𝑑𝑦 = 0$. Since $\frac{{\partial (2𝑥𝑦^3 + 𝑥^2)}}{\partial y}=6xy^2=\frac{{\partial (3𝑥^2𝑦^2 + 𝑦)}}{\partial x},$ we conclude that the differential equation is exact. Then $\frac{{\partial U}}{\partial x}=2𝑥𝑦^3 + 𝑥^2, \frac{{\partial U}}{\partial y}=3𝑥^2𝑦^2 + 𝑦.$ It follows that $U=x^2y^3+\frac{x^3}3+C(y),$ and hence $\frac{{\partial U}}{\partial y}=3x^2y^2+C'(y)=3x^2y^2+y.$ Therefore, $C'(y)=y,$ and thus $C(y)=\frac{y^2}2+C.$ We conclude that the general solution of the differential equation is $x^2y^3+\frac{x^3}3+\frac{y^2}2=C.$

3. Consider the equation $(𝑒^{2y} + 𝑥)𝑑𝑥 + (2𝑥𝑒^{2y} + 1)𝑑𝑦 = 0$. Since $\frac{{\partial (𝑒^{2y} + 𝑥)}}{\partial y}=2e^{2y}=\frac{{\partial (2𝑥𝑒^{2y} + 1)}}{\partial x},$ we conclude that the differential equation is exact. Then $\frac{{\partial U}}{\partial x}=𝑒^{2y} + 𝑥, \frac{{\partial U}}{\partial y}=2𝑥𝑒^{2y} + 1.$ It follows that $U=xe^{2y}+\frac{x^2}2+C(y),$ and hence $\frac{{\partial U}}{\partial y}=2xe^{2y}+C'(y)=2xe^{2y}+1.$ Therefore, $C'(y)=1,$ and thus $C(y)=y+C.$ We conclude that the general solution of the differential equation is $xe^{2y}+\frac{x^2}2+y=C.$