2.5 Exact Equations

In this section it’s convenient to write first order differential equations in the form

\begin{equation} M(x,y)\,dx+N(x,y)\,dy=0. \tag{2.5.1}\end{equation}

This equation can be interpreted as

\begin{equation} M(x,y)+N(x,y)\,{dy\over dx}=0, \tag{2.5.2}\end{equation}

where \(x\) is the independent variable and \(y\) is the dependent variable, or as

\begin{equation} M(x,y)\,{dx\over dy}+N(x,y)=0, \tag{2.5.3}\end{equation}

where \(y\) is the independent variable and \(x\) is the dependent variable. Since the solutions of (2.5.2) and (2.5.3) will often have to be left in implicit, form we’ll say that \(F(x,y)=c\) is an implicit solution of (2.5.1) if every differentiable function \(y=y(x)\) that satisfies \(F(x,y)=c\) is a solution of (2.5.2) and every differentiable function \(x=x(y)\) that satisfies \(F(x,y)=c\) is a solution of (2.5.3).

Here are some examples:

Equation (2.5.1)Equation (2.5.2)Equation (2.5.3)
\(3x^2y^2\,dx+2x^3y\,dy =0\)\(3x^2y^2+2x^3y\,\dst{dy\over dx} =0\)\(3x^2y^2\,\dst{dx\over dy}+2x^3y=0\)
\((x^2+y^2)\,dx +2xy\,dy=0\)\((x^2+y^2)+2xy\,\dst{dy\over dx}=0\)\((x^2+y^2)\,\dst{dx\over dy} +2xy=0\)
\(3y\sin x\,dx-2xy\cos x\,dy =0\)\(3y\sin x-2xy\cos x\,\dst{dy\over dx} =0\)\(3y\sin x\,\dst{dx\over dy}-2xy\cos x =0\)

Note that a separable equation can be written as (2.5.1) as

\[ M(x)\,dx+N(y)\,dy=0. \]

We’ll develop a method for solving (2.5.1) under appropriate assumptions on \(M\) and \(N\). This method is an extension of the method of separation of variables (Exercise 41). Before stating it we consider an example.

Example 2.5.1

Show that

\begin{equation} x^4y^3+x^2y^5+2xy=c \tag{2.5.4}\end{equation}

is an implicit solution of

\begin{equation} (4x^3y^3+2xy^5+2y)\,dx+(3x^4y^2+5x^2y^4+2x)\,dy=0. \tag{2.5.5}\end{equation}

Solution Regarding \(y\) as a function of \(x\) and differentiating (2.5.4) implicitly with respect to \(x\) yields

\[ (4x^3y^3+2xy^5+2y)+(3x^4y^2+5x^2y^4+2x)\,{dy\over dx}=0. \]

Similarly, regarding \(x\) as a function of \(y\) and differentiating (2.5.4) implicitly with respect to \(y\) yields

\[ (4x^3y^3+2xy^5+2y){dx\over dy}+(3x^4y^2+5x^2y^4+2x)=0. \]

Therefore (2.5.4) is an implicit solution of (2.5.5) in either of its two possible interpretations. ∎

You may think this example is pointless, since concocting a differential equation that has a given implicit solution isn’t particularly interesting. However, it illustrates the next important theorem, which we’ll prove by using implicit differentiation, as in Example 2.5.1.

Theorem 2.5.1

If \(F=F(x,y)\) has continuous partial derivatives \(F_x\) and \(F_y\), then

\begin{equation} F(x,y)=c\qquad \mbox{{\rm(}$c$=constant\mbox{)}}, \tag{2.5.6}\end{equation}

is an implicit solution of the differential equation

\begin{equation} F_x(x,y)\,dx+F_y(x,y)\,dy=0. \tag{2.5.7}\end{equation}

Proof Regarding \(y\) as a function of \(x\) and differentiating (2.5.6) implicitly with respect to \(x\) yields

\[ F_x(x,y)+F_y(x,y)\,{dy\over dx}=0. \]

On the other hand, regarding \(x\) as a function of \(y\) and differentiating (2.5.6) implicitly with respect to \(y\) yields

\[ F_x(x,y)\,{dx\over dy}+F_y(x,y)=0. \]

Thus, (2.5.6) is an implicit solution of (2.5.7) in either of its two possible interpretations. ∎

We’ll say that the equation

\begin{equation} M(x,y)\,dx+N(x,y)\,dy=0 \tag{2.5.8}\end{equation}

is exact on an an open rectangle \(R\) if there’s a function \(F=F(x,y)\) such \(F_x\) and \(F_y\) are continuous, and

\begin{equation} F_x(x,y)=M(x,y) \mbox{\quad and \quad} F_y(x,y)=N(x,y) \tag{2.5.9}\end{equation}

for all \((x,y)\) in \(R\). This usage of “exact” is related to its usage in calculus, where the expression

\[ F_x(x,y)\,dx+F_y(x,y)\,dy \]

(obtained by substituting (2.5.9) into the left side of (2.5.8)) is the exact differential of \(F\).

Example 2.5.1 shows that it’s easy to solve (2.5.8) if it’s exact and we know a function \(F\) that satisfies (2.5.9). The important questions are:

Question 1. Given an equation (2.5.8), how can we determine whether it’s exact?

Question 2. If (2.5.8) is exact, how do we find a function \(F\) satisfying (2.5.9)?

To discover the answer to Question 1, assume that there’s a function \(F\) that satisfies (2.5.9) on some open rectangle \(R\), and in addition that \(F\) has continuous mixed partial derivatives \(F_{xy}\) and \(F_{yx}\). Then a theorem from calculus implies that

\begin{equation} F_{xy}=F_{yx}. \tag{2.5.10}\end{equation}

If \(F_x=M\) and \(F_y=N\), differentiating the first of these equations with respect to \(y\) and the second with respect to \(x\) yields

\begin{equation} F_{xy}=M_y\mbox{\quad and \quad} F_{yx}=N_x. \tag{2.5.11}\end{equation}

From (2.5.10) and (2.5.11), we conclude that a necessary condition for exactness is that \(M_y=N_x\). This motivates the next theorem, which we state without proof.

Theorem 2.5.2

Suppose \(M\) and \(N\) are continuous and have continuous partial derivatives \(M_y\) and \(N_x\) on an open rectangle \(R.\) Then

\[ M(x,y)\,dx+N(x,y)\,dy=0 \]

is exact on \(R\) if and only if

\begin{equation} M_y(x,y)=N_x(x,y) \tag{2.5.12}\end{equation}

for all \((x,y)\) in \(R.\).

To help you remember the exactness condition, observe that the coefficients of \(dx\) and \(dy\) are differentiated in (2.5.12) with respect to the “opposite” variables; that is, the coefficient of \(dx\) is differentiated with respect to \(y\), while the coefficient of \(dy\) is differentiated with respect to \(x\).

Example 2.5.2

Show that the equation

\[ 3x^2y\,dx+4x^3\,dy=0 \]

is not exact on any open rectangle.

Solution Here

\[ M(x,y)=3x^2y\mbox{\quad and \quad} N(x,y)=4x^3 \]

so

\[ M_y(x,y)=3x^2 \mbox{\quad and \quad} N_x(x,y)=12 x^2. \]

Therefore \(M_y=N_x\) on the line \(x=0\), but not on any open rectangle, so there’s no function \(F\) such that \(F_x(x,y)=M(x,y)\) and \(F_y(x,y)=N(x,y)\) for all \((x,y)\) on any open rectangle. ∎

The next example illustrates two possible methods for finding a function \(F\) that satisfies the condition \(F_x=M\) and \(F_y=N\) if \(M\,dx+N\,dy=0 \) is exact.

Example 2.5.3

Solve

\begin{equation} (4x^3y^3+3x^2)\,dx+(3x^4y^2+6y^2)\,dy=0. \tag{2.5.13}\end{equation}

Solution (Method 1) Here

\[ M(x,y)=4x^3y^3+3x^2,\quad N(x,y)=3x^4y^2+6y^2, \]

and

\[ M_y(x,y)=N_x(x,y)=12 x^3y^2 \]

for all \((x,y)\). Therefore Theorem 2.5.2 implies that there’s a function \(F\) such that

\begin{equation} F_x(x,y)=M(x,y)=4x^3y^3+3x^2 \tag{2.5.14}\end{equation}

and

\begin{equation} F_y(x,y)=N(x,y)=3x^4y^2+6y^2 \tag{2.5.15}\end{equation}

for all \((x,y)\). To find \(F\), we integrate (2.5.14) with respect to \(x\) to obtain

\begin{equation} F(x,y)=x^4y^3+x^3+\phi(y), \tag{2.5.16}\end{equation}

where \(\phi (y)\) is the “constant” of integration. (Here \(\phi\) is “constant” in that it’s independent of \(x\), the variable of integration.) If \(\phi\) is any differentiable function of \(y\) then \(F\) satisfies (2.5.14). To determine \(\phi\) so that \(F\) also satisfies (2.5.15), assume that \(\phi\) is differentiable and differentiate \(F\) with respect to \(y\). This yields

\[ F_y(x,y)=3x^4y^2+\phi'(y). \]

Comparing this with (2.5.15) shows that

\[ \phi'(y)=6y^2. \]

We integrate this with respect to \(y\) and take the constant of integration to be zero because we’re interested only in finding some \(F\) that satisfies (2.5.14) and (2.5.15). This yields

\[ \phi (y)=2y^3. \]

Substituting this into (2.5.16) yields

\begin{equation} F(x,y)=x^4y^3+x^3+2y^3. \tag{2.5.17}\end{equation}

Now Theorem 2.5.1 implies that

\[ x^4y^3+x^3+2y^3=c \]

is an implicit solution of (2.5.13). Solving this for \(y\) yields the explicit solution

\[ y=\left(c-x^3\over2+x^4\right)^{1/3}. \]

Solution (Method 2) Instead of first integrating (2.5.14) with respect to \(x\), we could begin by integrating (2.5.15) with respect to \(y\) to obtain

\begin{equation} F(x,y)=x^4y^3+2y^3+\psi (x), \tag{2.5.18}\end{equation}

where \(\psi\) is an arbitrary function of \(x\). To determine \(\psi\), we assume that \(\psi\) is differentiable and differentiate \(F\) with respect to \(x\), which yields

\[ F_x(x,y)=4x^3y^3+\psi'(x). \]

Comparing this with (2.5.14) shows that

\[ \psi'(x)=3x^2. \]

Integrating this and again taking the constant of integration to be zero yields

\[ \psi(x)=x^3. \]

Substituting this into (2.5.18) yields (2.5.17).

Figure 2.5.1 shows a direction field and some integral curves of (2.5.13),

A direction field and integral curves for (4x^3y^3+3x^2)\,dx+(3x^4y^2+6y^2)\,dy=0
Figure 2.5.1. A direction field and integral curves for \((4x^3y^3+3x^2)\,dx+(3x^4y^2+6y^2)\,dy=0\)

Here’s a summary of the procedure used in Method 1 of this example. You should summarize procedure used in Method 2.

Procedure For Solving An Exact Equation

It’s a common mistake to omit Step 6. However, it’s important to include this step, since \(F\) isn’t itself a solution of (2.5.19).

Many equations can be conveniently solved by either of the two methods used in Example 2.5.3. However, sometimes the integration required in one approach is more difficult than in the other. In such cases we choose the approach that requires the easier integration.

Example 2.5.4

Solve the equation

\begin{equation} (ye^{xy} \tan x+e^{xy} \sec^2x)\,dx+xe^{xy} \tan x\,dy=0. \tag{2.5.21}\end{equation}

Solution We leave it to you to check that \(M_y=N_x\) on any open rectangle where \(\tan x\) and \(\sec x\) are defined. Here we must find a function \(F\) such that

\begin{equation} F_x(x,y)=ye^{xy} \tan x+e^{xy} \sec^2 x \tag{2.5.22}\end{equation}

and

\begin{equation} F_y(x,y)=xe^{xy} \tan x. \tag{2.5.23}\end{equation}

It’s difficult to integrate (2.5.22) with respect to \(x\), but easy to integrate (2.5.23) with respect to \(y\). This yields

\begin{equation} F(x,y)=e^{xy} \tan x+\psi (x). \tag{2.5.24}\end{equation}

Differentiating this with respect to \(x\) yields

\[ F_x(x,y)=ye^{xy}\tan x+e^{xy}\sec^2x+\psi'(x). \]

Comparing this with (2.5.22) shows that \(\psi'(x)=0\). Hence, \(\psi\) is a constant, which we can take to be zero in (2.5.24), and

\[ e^{xy} \tan x=c \]

is an implicit solution of (2.5.21). ∎

Attempting to apply our procedure to an equation that isn’t exact will lead to failure in Step 4, since the function

\[ N-{\partial G\over\partial y} \]

won’t be independent of \(x\) if \(M_y\ne N_x\) (Exercise 31), and therefore can’t be the derivative of a function of \(y\) alone. Here’s an example that illustrates this.

Example 2.5.5

Verify that the equation

\begin{equation} 3x^2y^2\,dx+6x^3y\,dy=0 \tag{2.5.25}\end{equation}

is not exact, and show that the procedure for solving exact equations fails when applied to (2.5.25).

Solution Here

\[ M_y(x,y)=6x^2y \mbox{\quad and \quad } N_x(x,y)=18x^2y, \]

so (2.5.25) isn’t exact. Nevertheless, let’s try to find a function \(F\) such that

\begin{equation} F_x(x,y)=3x^2y^2 \tag{2.5.26}\end{equation}

and

\begin{equation} F_y(x,y)=6x^3y. \tag{2.5.27}\end{equation}

Integrating (2.5.26) with respect to \(x\) yields

\[ F(x,y)=x^3y^2+\phi(y), \]

and differentiating this with respect to \(y\) yields

\[ F_y(x,y)=2x^3y+\phi'(y). \]

For this equation to be consistent with (2.5.27),

\[ 6x^3y=2x^3y+\phi'(y), \]

or

\[ \phi'(y)=4x^3y. \]

This is a contradiction, since \(\phi'\) must be independent of \(x\). Therefore the procedure fails.

2.5 Exercises

In Exercises 1–17 determine which equations are exact and solve them.

  1. \(6x^2y^2\,dx+4x^3y\,dy=0\)

    Show answer

    \(2x^3y^2=c\)

  2. \((3y\cos x+4xe^x+2x^2e^x)\,dx+(3\sin x+3)\,dy=0\)

    Show answer

    \(3y\sin x+2x^2e^x+3y=c\)

  3. \(14x^2y^3\,dx+21 x^2y^2\,dy=0\)

    Show answer

    Not exact

  4. \((2x-2y^2)\,dx+(12y^2-4xy)\,dy=0\)

    Show answer

    \(x^2-2xy^2+4y^3=c\)

  5. \((x+y)^2\,dx+(x+y)^2\,dy=0\)

    Show answer

    \(x+y=c\)

  6. \((4x+7y)\,dx+(3x+4y)\,dy=0\)

    Show answer

    Not exact

  7. \((-2y^2\sin x+3y^3-2x)\,dx+(4y\cos x+9xy^2)\,dy=0\)

    Show answer

    \(2y^2\cos x+3xy^3-x^2=c\)

  8. \((2x+y)\,dx+(2y+2x)\,dy=0\)

    Show answer

    Not exact

  9. \((3x^2+2xy+4y^2)\,dx+(x^2+8xy+18y)\,dy=0\)

    Show answer

    \(x^3+x^2y+4xy^2+9y^2=c\)

  10. \((2x^2+8xy+y^2)\,dx+(2x^2+xy^3/3)\,dy=0\)

    Show answer

    Not exact

  11. \(\dst{\left({1\over x}+2x\right)\,dx+ \left({1\over y}+2y\right)\,dy=0}\)

    Show answer

    \(\ln |xy|+x^2+y^2=c\)

  12. \((y\sin xy+xy^2\cos xy)\,dx+(x\sin xy+xy^2\cos xy)\,dy=0\)

    Show answer

    Not exact

  13. \(\dst{{x\,dx\over(x^2+y^2)^{3/2}}+{y\,dy \over(x^2+y^2)^{3/2}}=0}\)

    Show answer

    \(x^2+y^2=c\)

  14. \(\left(e^x(x^2y^2+2xy^2)+6x\right)\,dx+(2x^2ye^x+2)\,dy=0\)

    Show answer

    \(x^2y^2e^x+2y+3x^2=c\)

  15. \(\left(x^2e^{x^2+y}(2x^2+3)+4x\right)\,dx+(x^3e^{x^2+y}-12y^2)\,dy=0\)

    Show answer

    \(x^3e^{x^2+y}-4y^3+2x^2=c\)

  16. \(\left(e^{xy}(x^4y+4x^3)+3y\right)\,dx+(x^5e^{xy}+3x)\,dy=0\)

    Show answer

    \(x^4e^{xy}+3xy=c\)

  17. \((3x^2\cos xy-x^3y\sin xy+4x)\,dx+(8y-x^4\sin xy)\,dy=0\)

    Show answer

    \(x^3\cos xy+4y^2+2x^2=c\)

In Exercises 18–22 solve the initial value problem.

  1. \((4x^3y^2-6x^2y-2x-3)\,dx+(2x^4y-2x^3)\,dy=0,\quad y(1)=3\)

    Show answer

    \(y=\dst{x+\sqrt{2x^2+3x-1}\over x^2}\)

  2. \((-4y\cos x+4\sin x\cos x+\sec^2x)\,dx+ (4y-4\sin x)\,dy=0,\quad y(\pi/4)=0\)

    Show answer

    \(y=\sin x-\dst{\sqrt{1-{\tan x\over 2}}}\)

  3. \((y^3-1)e^x\,dx+3y^2(e^x+1)\,dy=0,\quad y(0)=0\)

    Show answer

    \(\dst{y=\left({e^x-1\over e^x+1}\right)^{1/3}}\)

  4. \((\sin x-y\sin x-2\cos x)\,dx+\cos x\,dy=0,\quad y(0)=1\)

    Show answer

    \(y=1+2\tan x\)

  5. \((2x-1)(y-1)\,dx+(x+2)(x-3)\,dy=0,\quad y(1)=-1\)

    Show answer

    \(y=\dst{x^2-x+6\over(x+2)(x-3)}\)

  6. C/G Solve the exact equation

    \[ (7x+4y)\,dx+(4x+3y)\,dy=0. \]

    Plot a direction field and some integral curves for this equation on the rectangle

    \[ \{-1\le x\le1,-1\le y\le1\}. \]
    Show answer

    \(\dst{{7x^2\over2}+4xy+{3y^2\over 2}=c}\)

  7. C/G Solve the exact equation

    \[ e^x(x^4y^2+4x^3y^2+1)\,dx+(2x^4ye^x+2y)\,dy=0. \]

    Plot a direction field and some integral curves for this equation on the rectangle

    \[ \{-2\le x\le2,-1\le y\le1\}. \]
    Show answer

    \((x^4y^2+1)e^x+y^2=c\)

  8. C/G Plot a direction field and some integral curves for the exact equation

    \[ (x^3y^4+x)\,dx+(x^4y^3+y)\,dy=0 \]

    on the rectangle \(\{-1\le x\le 1,-1\le y\le1\}\). (See Exercise 37(a)).

  9. C/G Plot a direction field and some integral curves for the exact equation

    \[ (3x^2+2y)\,dx+(2y+2x)\,dy=0 \]

    on the rectangle \(\{-2\le x\le 2,-2\le y\le2\}\). (See Exercise 37(b)).

  10. L

    1. Solve the exact equation

      \[ (x^3y^4+2x)\,dx+(x^4y^3+3y)\,dy=0 \tag*{\rm(A)} \]

      implicitly.

    2. For what choices of \((x_0,y_0)\) does Theorem 2.3.1 imply that the initial value problem

      \[ (x^3y^4+2x)\,dx+(x^4y^3+3y)\,dy=0,\quad y(x_0)=y_0, \tag*{\rm(B)} \]

      has a unique solution on an open interval \((a,b)\) that contains \(x_0\)?

    3. Plot a direction field and some integral curves for (A) on a rectangular region centered at the origin. What is the interval of validity of the solution of (B)?

  11. L

    1. Solve the exact equation

      \[ (x^2+y^2)\,dx+2xy\,dy=0 \tag*{\rm(A)} \]

      implicitly.

    2. For what choices of \((x_0,y_0)\) does Theorem 2.3.1 imply that the initial value problem

      \[ (x^2+y^2)\,dx+2xy\,dy=0,\quad y(x_0)=y_0, \tag*{\rm(B)} \]

      has a unique solution \(y=y(x)\) on some open interval \((a,b)\) that contains \(x_0\)?

    3. Plot a direction field and some integral curves for (A). From the plot determine, the interval \((a,b)\) of (b), the monotonicity properties (if any) of the solution of (B), and \(\lim_{x\to a+}y(x)\) and \(\lim_{x\to b-}y(x)\). Hint: Your answers will depend upon which quadrant contains \((x_0,y_0)\).

  12. Find all functions \(M\) such that the equation is exact.

    1. \(M(x,y)\,dx+(x^2-y^2)\,dy=0\)

    2. \(M(x,y)\,dx+2xy\sin x\cos y\,dy=0\)

    3. \(M(x,y)\,dx+(e^x-e^y\sin x)\,dy=0\)

    Show answer

    (a) \(M(x,y)=2xy+f(x)\) (b) \(M(x,y)=2(\sin x+x\cos x) (y\sin y+\cos y)+f(x)\)

       (c) \(M(x,y)=ye^x-e^y\cos x+f(x)\)

  13. Find all functions \(N\) such that the equation is exact.

    1. \((x^3y^2+2xy+3y^2)\,dx+N(x,y)\,dy=0\)

    2. \((\ln xy+2y\sin x)\,dx+N(x,y)\,dy=0\)

    3. \((x\sin x+y\sin y)\,dx+N(x,y)\,dy=0\)

    Show answer

    (a) \(N(x,y)=\dst{x^4y\over 2}+x^2+6xy+g(y)\) (b) \(N(x,y)=\dst{x\over y}+2y\sin x+g(y)\)

       (c) \(N(x,y)=x(\sin y+y\cos y)+g(y)\)

  14. Suppose \(M,N,\) and their partial derivatives are continuous on an open rectangle \(R\), and \(G\) is an antiderivative of \(M\) with respect to \(x\); that is,

    \[ {\partial G\over\partial x}=M. \]

    Show that if \(M_y\ne N_x\) in \(R\) then the function

    \[ N-{\partial G\over\partial y} \]

    is not independent of \(x\).

  15. Prove: If the equations \(M_1\,dx+N_1\,dy=0\) and \(M_2\, dx+N_2\,dy=0\) are exact on an open rectangle \(R\), so is the equation

    \[(M_1+M_2)\,dx+(N_1+N_2)\,dy=0.\]

  16. Find conditions on the constants \(A\), \(B\), \(C\), and \(D\) such that the equation

    \[ (Ax+By)\,dx+(Cx+Dy)\,dy=0 \]

    is exact.

    Show answer

    \(B=C\)

  17. Find conditions on the constants \(A\), \(B\), \(C\), \(D\), \(E\), and \(F\) such that the equation

    \[ (Ax^2+Bxy+Cy^2)\,dx+(Dx^2+Exy+Fy^2)\,dy=0 \]

    is exact.

    Show answer

    \(B=2D,\quad E=2C\)

  18. Suppose \(M\) and \(N\) are continuous and have continuous partial derivatives \(M_y\) and \(N_x\) that satisfy the exactness condition \(M_y=N_x\) on an open rectangle \(R\). Show that if \((x,y)\) is in \(R\) and

    \[ F(x,y)=\int^x_{x_0}M(s,y_0)\,ds+\int^y_{y_0}N(x,t)\,dt, \]

    then \(F_x=M\) and \(F_y=N\).

  19. Under the assumptions of Exercise 35, show that

    \[ F(x,y)=\int^y_{y_0}N(x_0,s)\,ds+\int^x_{x_0}M(t,y)\,dt. \]

  20. Use the method suggested by Exercise 35, with \((x_0,y_0)=(0,0)\), to solve the these exact equations:

    1. \((x^3y^4+x)\,dx+(x^4y^3+y)\,dy=0\)

    2. \((x^2+y^2)\,dx+2xy\,dy=0\)

    3. \((3x^2+2y)\,dx+(2y+2x)\,dy=0\)

    Show answer

    (a) \(2x^2+x^4y^4+y^2=c\) (b) \(x^3+3xy^2=c\) (c) \(x^3+y^2+2xy=c\)

  21. Solve the initial value problem

    \[ y'+{2\over x}y=-{2xy\over x^2+2x^2y+1},\quad y(1)=-2. \]
    Show answer

    \(y=-1-\dst{1\over x^2}\)

  22. Solve the initial value problem

    \[ y'-{3\over x}y={2x^4(4x^3-3y)\over3x^5+3x^3+2y},\quad y(1)=1. \]
    Show answer

    \(y=x^3\dst{\left(-3(x^2+1)+\sqrt{9x^4+34x^2+21}\over2\right)}\)

  23. Solve the initial value problem

    \[ y'+2xy=-e^{-x^2}\left({3x+2ye^{x^2}\over2x+3ye^{x^2}}\right),\quad y(0)=-1. \]
    Show answer

    \(y=-e^{-x^2}\dst{\left(2x+\sqrt{9-5x^2}\over3\right)}\).

  24. Rewrite the separable equation

    \[ h(y)y'=g(x) \tag*{\rm (A)} \]

    as an exact equation

    \[ M(x,y)\,dx+N(x,y)\,dy=0. \tag*{\rm (B)} \]

    Show that applying the method of this section to (B) yields the same solutions that would be obtained by applying the method of separation of variables to (A)

  25. Suppose all second partial derivatives of \(M=M(x,y)\) and \(N=N(x,y)\) are continuous and \(M\,dx+N\,dy=0\) and \(-N\,dx+M\,dy=0\) are exact on an open rectangle \(R\). Show that \(M_{xx}+M_{yy}=N_{xx}+N_{yy}=0\) on \(R\).

  26. Suppose all second partial derivatives of \(F=F(x,y)\) are continuous and \(F_{xx}+F_{yy}=0\) on an open rectangle \(R\). (A function with these properties is said to be harmonic; see also Exercise 42.) Show that \(-F_y\,dx+F_x\,dy=0\) is exact on \(R\), and therefore there’s a function \(G\) such that \(G_x=-F_y\) and \(G_y=F_x\) in \(R\). (A function \(G\) with this property is said to be a harmonic conjugate of \(F\).)

  27. Verify that the following functions are harmonic, and find all their harmonic conjugates. (See Exercise 43.)

    (a) \(x^2-y^2\)(b) \(e^x\cos y\)(c) \(x^3-3xy^2\)
    (d) \(\cos x\cosh y\)(e) \(\sin x\cosh y\)
    Show answer

    (a) \(G(x,y)=2xy+c\) (b) \(G(x,y)=e^x\sin y+c\)

    (c) \(G(x,y)=3x^2y-y^3+c\) (d) \(G(x,y)=-\sin x\sinh y+c\)

    (e) \(G(x,y)=\cos x\sinh y+c\)