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Mathematics LibreTexts

2.7: Exact Differential Equations

Consider the equation

\[        f(x,y)  =  C \]

Taking the gradient we get

\[       f_x(x,y)\hat{\textbf{i}} + f_y(x,y)\hat{\textbf{j}}  =  0\]

We can write this equation in differential form as 

  \[      f_x(x,y)\, dx+ f_y(x,y)\, dy  =  0\]

Now divide by \( dx \) (we are not pretending to be rigorous here) to get

\[        f_x(x,y)+ f_y(x,y) \dfrac{dy}{dx}  =  0\]

Which is a first order differential equation. The goal of this section is to go backward. That is if a differential equation if of the form above, we seek the original function \(f(x,y)\) (called a potential function). A differential equation with a potential function is called exact. If you have had vector calculus, this is the same as finding the potential functions and using the fundamental theorem of line integrals.

Example 1

Solve

\[ 4xy + 1 + (2x^2 + \cos y)y'  =  0 \]

Solution

We seek a function \(f(x,y)\) with 

\[  f_x(x,y)  =  4xy + 1 \;\;\; \text{and}  f_y(x,y)  =  2x^2 + \cos y \]

Integrate the first equation with respect to \(x\) to get

\[ f(x,y)  =  2x^2y + x + C(y)   \]

Notice since \(y\) is treated as a constant, we write \(C(y)\). Now take the partial derivative with respect to y to get

\[  f_y(x,y)  =  2x^2 + C'(y) \]

We have two formulae for \( f_y(x,y) \) so we can set them equal to eachother.

\[ 2x^2 + \cos y  =  2x^2 + C'(y) \]

That is 

\[ C'(y)  =  \cos\,  y \]

or

\[  C(y)  =  \sin \, y \]

Hence 

\[  f(x,y)  =  2x^2y + x + \sin \, y  \]

The solution to the differential equation is

\[ 2x^2y + x + \sin \, y  =  C \]

Does this method always work? The answer is no. We can tell if the method works by remembering that for a function with continuous partial derivatives, the mixed partials are order independent. That is 

\[   f_{xy}  =  f_{yx} \]

If we have the differential equation

\[   M(x,y) + N(x,y)y'  =  0  \]

then we say it is an exact differential equation if 

\[   M_y(x,y)  =  N_x(x,y)  \]

Theorem (Solutions to Exact Differential Equations)

Let \(M\), \(N\), \(M_y\), and \(N_x\) be continuous with

\[  M_y  =  N_x\]

Then there is a function \(f(x,y)\) with 

\( f_x  =  M \)    and     \(   f_y  =  N \)

such that 

\[  f(x,y)  =  C \]

is a solution to the differential equation 

\[ M(x,y)  + N(x,y)y'  =  0 \]

Example 2

Solve the differential equation 

\[  y + (2xy - e^{-2y})y'  =  0  \]

Solution

We have 

\[  M(x,y)  =  y   \text{and}    N(x,y)  =  2xy - e^{-2y}   \]

Now calculate 

\(  M_y  =  1  \)   and   \( N_x  =  2y \)

Since they are not equal, finding a potential function \(f\) is hopeless. However there is a glimmer of hope if we remember how we solved first order linear differential equations. We multiplied both sides by an integrating factor \(m\). We do that here to get

\[   mM + mN_y'  =  0 \]

For this to be exact we must have 

\[ (mM)_y  =   (mN)_x  \]

Using the product rule gives

\[   m_yM + mM_y  =   m_xN + mN_x  \]

We now have a new differential equation that is unfortunately more difficult to solve than the original differential equation. We simplify the equation by assuming that either m is a function of only \(x\) or only \(y\). If it is a function of only \(x\), then \( m_y = 0 \) and 

\[  mM_y  =   m_xN + mN_x  \]

Solving for \(m_x\), we get

\[ m_x = \dfrac{M_y-N_x}{N} \]

If this is a function of \(y\) only, then we will be able to find an integrating factor that involves \(y\) only. If it is a function of only \(y\), then \( m_x  =  0\) and 

\[  m_yM + mM_y  =  mN_x   \]

Solving for \(m_y\), we get

\[ m_y = \dfrac{N_x-M_y}{M} m \]

If this is a function of \(y\) only, then we will be able to find an integrating factor that involves \(y\) only.

For our example

\[ m_y  = \dfrac{N_x - M_y  }{M} m = \dfrac{2y-1}{y} m = (2-\frac{1}{y})m \]

Separating gives

\[ \dfrac{dm}{m} = (2-\frac{1}{y}) \,dy \]

Integrating gives

     \[    ln \, m  =  2y  -  ln\, y \]

      \[  m  =  e^{2y - ln\, y}  =  y ^{-1}e^{2y} \]

Multiplying both sides of the original differential equation by \(m\) gives

\[        y(y ^{-1}e^{2y}) + (y ^{-1}e^{2y})(2xy - e^{-2y})y'  =  0  \]

\[   e^{2y} + (2xe^{2y} - \frac{1}{y})y'  =  0  \]

Now we see that 

\[   M_y  =  2e^{2y}  =   N_x  \]

Which tells us that the differential equation is exact.  We therefore have

\[   f_x (x,y)  =  e^{2y}  \]

Integrating with respect to \(x\) gives

\[  f(x,y)  =  xe^{2y} + C(y) \]

Now taking the partial derivative with respect to y gives

\[    f_y(x,y)  =  2xe^{2y} + C'(y)  =  2xe^{2y} - \frac{1}{y} \]

So that 

\[   C'(y)  =  \frac{1}{y} \]

Integrating gives

\[  C(y)  =  ln\, y  \]

The final solution is 

\[  xe^{2y} + ln\, y  =  0 \]

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