exact differential equation

Let $ R$ be a region in $ \mathbb{R}^2$ and let the functions $ X\!: R \to \mathbb{R}$ , $ Y\!: R \to \mathbb{R}$ have continuous partial derivatives in $ R$ . The first order differential equation

$\displaystyle X(x,\,y)+Y(x,\,y)\frac{dy}{dx} = 0$

or

$\displaystyle X(x,\,y)dx+Y(x,\,y)dy = 0$ (1)

is called an exact differential equation, if the condition

$\displaystyle \frac{\partial X}{\partial y} = \frac{\partial Y}{\partial x}$

is true in $ R$ .

Then there is a function $ f\!: R \to \mathbb{R}$ such that the equation (1) has the form

$\displaystyle d\,f(x,\,y) = 0,$

whence its general integral is

$\displaystyle f(x,\,y) = C.$

The solution function $ f$ can be calculated as the line integral

$\displaystyle f(x,\,y) := \int_{P_0}^P [X(x,\,y)\,dx+Y(x,\,y)\,dy]$ (2)

along any curve $ \gamma$ connecting an arbitrarily chosen point $ P_0 =(x_0,\,y_0)$ and the point $ P = (x,\,y)$ in the region $ R$ (the integrating factor is now $ \equiv 1$ ).

Example. Solve the differential equation

$\displaystyle \frac{2x}{y^3}\,dx+\frac{y^2-3x^2}{y^4}\,dy = 0.$

This equation is exact, since

$\displaystyle \frac{\partial}{\partial y}\frac{2x}{y^3} = -\frac{6x}{y^4} = \frac{\partial}{\partial x}\frac{y^2-3x^2}{y^4}.$

If we use as the integrating way the broken line from $ (0,\,1)$ to $ (x,\,1)$ and from this to $ (x,\,y)$ , the integral (2) is simply

$\displaystyle \int_0^x\frac{2x}{1^3}\,dx+\!\int_1^y\frac{y^2-3x^2}{y^4}\,dy = \...
...}{y}+1
= x^2-\frac{1}{y}+\frac{x^2}{y^3}+1-x^2 = \frac{x^2}{y^3}-\frac{1}{y}+1.$

Thus we have the general integral

$\displaystyle \frac{x^2}{y^3}-\frac{1}{y} = C$

of the given differential equation.



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As of this snapshot date, this entry was owned by pahio.