Exact equations

Consider the following differential equation:\[3x^2 + 2xy \,dx + (x^2 + y^2) \,dy = 0\]To determine if it is exact, calculate the partial derivatives:\[\frac{\partial M}{\partial y} = \frac{\partial (3x^2 + 2xy)}{\partial y} = 2x\]\[\frac{\partial N}{\partial x} = \frac{\partial (x^2 + y^2)}{\partial x} = 2x\]Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the equation is exact.

Returning to our previous example, find a potential function \(\phi(x, y)\) such that:\[\frac{\partial \phi}{\partial x} = M(x, y) = 3x^2 + 2xy\]\[\frac{\partial \phi}{\partial y} = N(x, y) = x^2 + y^2\]We integrate \(\frac{\partial \phi}{\partial x}\) with respect to \(x\):\[\phi(x, y) = \int (3x^2 + 2xy) \,dx = x^3 + x^2y + f(y)\]Now, differentiate this with respect to \(y\) and compare with \(\frac{\partial \phi}{\partial y}\):\[\frac{\partial \phi}{\partial y} = x^2 + f'(y) = x^2 + y^2\]Thus, \(f'(y) = y^2\) and \(f(y) = \frac{y^3}{3}\). So the potential function is:\[\phi(x, y) = x^3 + x^2y + \frac{y^3}{3}\]

Consider the differential equation:\[2xy \,dx + (x^2 + 2y) \,dy = 0\]To check for exactness, calculate the partial derivatives:\[\frac{\partial M}{\partial y} = \frac{\partial (2xy)}{\partial y} = 2x\]\[\frac{\partial N}{\partial x} = \frac{\partial (x^2 + 2y)}{\partial x} = 2x\]Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the equation is exact.

Consider the equation from the last example:\[2xy \,dx + (x^2 + 2y) \,dy = 0\]Follow these steps:1. Integrate \(M(x, y) = 2xy\) with respect to \(x\):\[\phi(x, y) = \int 2xy \,dx = x^2y + h(y)\]2. Differentiate \(\phi(x, y)\) with respect to \(y\) and equate to \(N(x, y)\):\[\frac{\partial \phi}{\partial y} = x^2 + h'(y) = x^2 + 2y\]3. Solve for \(h'(y)\):\[h'(y) = 2y\] and \[h(y) = y^2\]4. Combine the results to obtain:\[\phi(x, y) = x^2y + y^2\]5. The solution to the differential equation is given by:\[\phi(x, y) = C\] or \[x^2y + y^2 = C\]

Consider the differential equation:\[2xy \,dx + (x^2 + y) \,dy = 0\]To determine if it is exact, calculate the partial derivatives:\[\frac{\partial M}{\partial y} = \frac{\partial (2xy)}{\partial y} = 2x\]\[\frac{\partial N}{\partial x} = \frac{\partial (x^2 + y)}{\partial x} = 2x\]Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the equation is exact.To find the solution, we need a potential function \(\phi(x, y)\):Integrate \(2xy\) with respect to \(x\):\[\phi(x, y) = \int 2xy \,dx = x^2y + h(y)\]Differentiate with respect to \(y\) and equate to \(N(x, y)\):\[\frac{\partial \phi}{\partial y} = x^2 + h'(y) = x^2 + y\]Thus,\[h'(y) = y\] and \[h(y) = \frac{y^2}{2}\]The potential function is:\[\phi(x, y) = x^2y + \frac{y^2}{2}\]The solution to the differential equation is:\[x^2y + \frac{y^2}{2} = C\]

Consider the differential equation:\[(3x^2y + y^3) \,dx + (x^3 + 3xy^2 + y) \,dy = 0\]To determine if it is exact, calculate the partial derivatives:\[\frac{\partial M}{\partial y} = \frac{\partial (3x^2y + y^3)}{\partial y} = 3x^2 + 3y^2\]\[\frac{\partial N}{\partial x} = \frac{\partial (x^3 + 3xy^2 + y)}{\partial x} = 3x^2 + 3y^2\]Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the equation is exact.Integrate \(3x^2y + y^3\) with respect to \(x\):\[\phi(x, y) = \int (3x^2y + y^3) \,dx = x^3y + xy^3 + h(y)\]Dfferentiate with respect to \(y\) and equate to \(N(x, y)\):\[\frac{\partial \phi}{\partial y} = x^3 + 3xy^2 + h'(y) = x^3 + 3xy^2 + y\]Thus,\[h'(y) = y\] and \[h(y) = \frac{y^2}{2}\]The potential function is:\[\phi(x, y) = x^3y + xy^3 + \frac{y^2}{2}\]The solution to the differential equation is:\[x^3y + xy^3 + \frac{y^2}{2} = C\]

Consider the differential equation:\[(2x + 3y^2) \,dx + (6xy - 4) \,dy = 0\]To determine if it is exact, calculate the partial derivatives:\[\frac{\partial M}{\partial y} = \frac{\partial (2x + 3y^2)}{\partial y} = 6y\]\[\frac{\partial N}{\partial x} = \frac{\partial (6xy - 4)}{\partial x} = 6y\]Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the equation is exact.To solve it, follow these steps:1. Integrate \(2x + 3y^2\) with respect to \(x\):\[\phi(x, y) = \int (2x + 3y^2) \,dx = x^2 + 3xy^2 + h(y)\]2. Differentiate with respect to \(y\) and equate to \(N(x, y)\):\[\frac{\partial \phi}{\partial y} = 6xy + h'(y) = 6xy - 4\]3. Solve for \(h'(y)\):\[h'(y) = -4\] and \[h(y) = -4y\]4. The potential function is:\[\phi(x, y) = x^2 + 3xy^2 - 4y\]5. The solution to the differential equation is:\[x^2 + 3xy^2 - 4y = C\]

Consider an engineering problem where the temperature distribution along a rod is described by the differential equation:\[k \frac{\text{d}T}{\text{d}x} + T \frac{\text{d}x}{\text{d}t} = 0\]To determine the temperature \(T\) along the rod over time, you identify it as an exact equation and employ the methods to solve it.

Consider the motion of a particle in a potential field described by the equation:\[F(x, y) \,dx + G(x, y) \,dy = 0\]By determining the potential function, you can find the path of the particle and understand its behaviour within the field.

Consider an economic model where the rate of change of an investment value is described by:\[rV \,dV + b \,dI = 0\]By solving this exact equation, you can determine how the investment changes over time and optimise economic outcomes.