Question #344033

Solve the equation (𝑥 + 𝑠𝑖𝑛𝑦)𝑑𝑥 + (𝑦2 + 𝑥𝑐𝑜𝑠𝑦)𝑑𝑦 = 0


1
Expert's answer
2022-05-24T13:48:28-0400
Qx=cosy,Py=cosy\dfrac{\partial Q}{\partial x}=\cos y,\dfrac{\partial P}{\partial y}=\cos y

We have the following system of differential equations to find the function  u(x,y)u(x, y)


u=(x+siny)dx=x22+xsiny+φ(y)u=\int(x+\sin y)dx=\dfrac{x^2}{2}+x\sin y+\varphi(y)

uy=xcosy+φ(y)\dfrac{\partial u}{\partial y}=x\cos y+\varphi'(y)

xcosy+φ(y)=y2+xcosyx\cos y+\varphi'(y)=y^2 +x\cos y

φ(y)=y2\varphi'(y)=y^2

φ(y)=y33C\varphi(y)=\dfrac{y^3}{3}-C

So that the general solution of the exact differential equation is given by


x22+xsiny+y33=C\dfrac{x^2}{2}+x\sin y+\dfrac{y^3}{3}=C



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