Exact differential equations

5.3. Exact differential equations#

5.3.1. Skill builder problems#

Obtain general solutions to:

  1. \(-2xy \sin( x^2) \d{x} + \cos(x^2) \d{y} = 0\)

    Solution

    The ODE is already in the standard form so

    (5.28)#\[\begin{align} P &= -2xy \sin(x^2) \\ Q &= \cos(x^2) \\ \end{align}\]

    Check to see if the ODE is exact:

    (5.29)#\[\begin{align} \td{}{P}{y}{x} &= -2x \sin(x^2) \\ \td{}{Q}{x}{y} &= -2x \sin(x^2) \end{align}\]

    The two partial derivatives are equal, so the ODE is exact. You can proceed directly to integration. First, integrate Q with respect y

    (5.30)#\[\begin{equation} f(x,y) = \int \cos(x^2) \d{y} = y \cos(x^2) + k(x) \end{equation}\]

    where k is an unknown function of x. Then, differentiate f with respect to x and compare to P:

    (5.31)#\[\begin{align} \td{}{f}{x}{y} = -2xy \sin(x^2) + k'(x) &= P = -2xy \sin(x^2) \\ k'(x) &= 0 \end{align}\]

    This simple ODE has \(k = 0\) as a solution (neglecting the integration constant). Putting it all together,

    (5.32)#\[\begin{equation} f = y \cos(x^2) = c \end{equation}\]

    is an implicit solution of the ODE, which we can manipulate to an explicit solution:

    (5.33)#\[\begin{equation} y = \frac{c}{\cos(x^2)} \end{equation}\]

  2. \(y' = \dfrac{x^4 + y^2}{xy}\)

    Solution

    This ODE is not in the standard form, so we need to first rearrange:

    (5.34)#\[\begin{align} xy \d{y} = (x^4+y^2) \d{x} \\ (x^4 + y^2) \d{x} - xy \d{y} = 0 \end{align}\]

    so:

    (5.35)#\[\begin{align} P &= x^4 + y^2 \\ Q &= -xy \end{align}\]

    Check to see if the ODE is exact:

    (5.36)#\[\begin{align} \td{}{P}{y}{x} &= 2y \\ \td{}{Q}{x}{y} &= -y \end{align}\]

    The two partial derivatives are not equal, so the ODE is not exact. In order to make it exact, we need to find an integrating factor F. First, compute:

    (5.37)#\[\begin{align} R &= \frac{1}{Q}\left[\td{}{P}{y}{x} - \td{}{Q}{x}{y} \right] \\ &=\frac{1}{-xy}(2y-(-y)) \\ &= -\frac{3}{x} \end{align}\]

    R is a function of only x, so use it to compute F

    (5.38)#\[\begin{equation} F = \exp\left(\int \frac{-3}{x} \d{x}\right) = e^{-3\ln(x)} = x^{-3} \end{equation}\]

    Apply the integrating factor to the original ODE:

    (5.39)#\[\begin{align} x^{-3}(x^4+y^2) \d{x} - x^{-3}(xy) \d{y} &= 0 \\ \left(x+\frac{y^2}{x^{-3}}\right) \d{x} - \frac{y}{x^2} \d{y} &= 0 \end{align}\]

    Integrate the Q of our exact ODE with respect to y:

    (5.40)#\[\begin{equation} f(x,y) = \int -\frac{y}{x^2} \d{y} = \frac{-y^2}{2x^2}+k(x) \end{equation}\]

    where k is an unknown function of x. Then, differentiate f with respect to x and compare to P of the exact ODE:

    (5.41)#\[\begin{align} \td{}{f}{x}{y} = \frac{y^2}{x^3} + k'(x) &= P = x + \frac{y^2}{x^3} \\ k'(x) &= x \\ \end{align}\]

    This ODE for k can be integrated directly (neglecting the integration constant)

    (5.42)#\[\begin{equation} k = \int x \d{x} = \frac{x^2}{2} \end{equation}\]

    Putting it all together,

    (5.43)#\[\begin{align} f = \frac{-y^2}{2x^2} + \frac{x^2}{2} = c \end{align}\]

    is an implicit solution of the ODE.

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