# Exact Differential Equations

## Solved Problems

### Example 3.

Solve the differential equation ${e^y}dx + \left( {2y + x{e^y}} \right)dy = 0.$

Solution.

First we check this equation for exactness:

$\frac{{\partial Q}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {2y + x{e^y}} \right) = {e^y},\;\; \frac{{\partial P}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {{e^y}} \right) = {e^y}.$

We see that $$\frac{{\partial Q}}{{\partial x}} = \frac{{\partial P}}{{\partial y}},$$ so that this equation is exact. Find the function $$u\left( {x,y} \right)$$ from the system of equations:

$\left\{ \begin{array}{l} \frac{{\partial u}}{{\partial x}} = {e^y}\\ \frac{{\partial u}}{{\partial y}} = 2y + x{e^y} \end{array} \right..$

Hence,

$u\left( {x,y} \right) = \int {P\left( {x,y} \right)dx} = \int {{e^y}dx} = x{e^y} + \varphi \left( y \right).$

Now, by differentiating this expression with respect to $$y$$ and equating it to $$\frac{{\partial u}}{{\partial y}},$$ we find the derivative $$\varphi'\left( y \right):$$

$\frac{{\partial u}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {x{e^y} + \varphi \left( y \right)} \right] = 2y + x{e^y},\;\; \Rightarrow \cancel{x{e^y}} + \varphi'\left( y \right) = 2y + \cancel{x{e^y}},\;\; \Rightarrow \varphi'\left( y \right) = 2y.$

As a result, we find $${\varphi \left( y \right)}$$ and the entire function $$u\left( {x,y} \right):$$

$\varphi \left( y \right) = \int {2ydy} = {y^2},\;\; \Rightarrow u\left( {x,y} \right) = x{e^y} + \varphi \left( y \right) = x{e^y} + {y^2}.$

Thus, the general solution of the differential equation is

$x{e^y} + {y^2} = C.$

### Example 4.

Solve the equation

$\left( {2xy - \sin x} \right)dx + \left( {{x^2} - \cos y} \right)dy = 0.$

Solution.

This differential equation is exact because

$\frac{{\partial Q}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {{x^2} - \cos y} \right) = 2x = \frac{{\partial P}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {2xy - \sin x} \right) = 2x.$

We find the function $$u\left( {x,y} \right)$$ from the system of two equations:

$\left\{ \begin{array}{l} \frac{{\partial u}}{{\partial x}} = 2xy - \sin x\\ \frac{{\partial u}}{{\partial y}} = {x^2} - \cos y \end{array} \right..$

By integrating the 1st equation with respect to the variable $$x,$$ we have

$u\left( {x,y} \right) = \int {\left( {2xy - \sin x} \right)dx} = {x^2}y + \cos x + \varphi \left( y \right).$

Plugging in the $$2$$nd equation, we obtain

$\frac{{\partial u}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {{x^2}y + \cos x + \varphi \left( y \right)} \right] = {x^2} - \cos y,\;\; \Rightarrow \cancel{x^2} + \varphi'\left( y \right) = \cancel{x^2} - \cos y,\;\; \Rightarrow \varphi'\left( y \right) = - \cos y.$

Hence,

$\varphi \left( y \right) = \int {\left( { - \cos y} \right)dy} = - \sin y.$

Thus, the function $$u\left( {x,y} \right)$$ is

$u\left( {x,y} \right) = {x^2}y + \cos x - \sin y,$

so that the general solution of the differential equation is given by the implicit formula:

${x^2}y + \cos x - \sin y = C.$

### Example 5.

Solve the equation

$\left( {1 + 2x\sqrt {{x^2} - {y^2}} } \right)dx - 2y\sqrt {{x^2} - {y^2}} dy = 0.$

Solution.

First of all we determine whether this equation is exact:

$\frac{{\partial Q}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( { - 2y\sqrt {{x^2} - {y^2}} } \right) = - 2y \cdot \frac{{2x}}{{2\sqrt {{x^2} - {y^2}} }} = - \frac{{2xy}}{{\sqrt {{x^2} - {y^2}} }},$
$\frac{{\partial P}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {1 + 2x\sqrt {{x^2} - {y^2}} } \right) = 2x \cdot \frac{{\left( { - 2y} \right)}}{{2\sqrt {{x^2} - {y^2}} }} = - \frac{{2xy}}{{\sqrt {{x^2} - {y^2}} }}.$

As you can see, $$\frac{{\partial Q}}{{\partial x}} = \frac{{\partial P}}{{\partial y}}.$$ Hence, this equation is exact. Find the function $$u\left( {x,y} \right),$$ satisfying the system of equations:

$\left\{ \begin{array}{l} \frac{{\partial u}}{{\partial x}} = 1 + 2x\sqrt {{x^2} - {y^2}} \\ \frac{{\partial u}}{{\partial y}} = - 2y\sqrt {{x^2} - {y^2}} \end{array} \right..$

Integrating the first equation gives:

$u\left( {x,y} \right) = \int {\left( {1 + 2x\sqrt {{x^2} - {y^2}} } \right)dx} = x + \frac{{{{\left( {{x^2} - {y^2}} \right)}^{\frac{3}{2}}}}}{{\frac{3}{2}}} + \varphi \left( y \right) = x + \frac{2}{3}{\left( {{x^2} - {y^2}} \right)^{\frac{3}{2}}} + \varphi \left( y \right),$

where $$\varphi \left( y \right)$$ is a certain unknown function of $$y$$ that will be defined later.

We substitute the result into the second equation of the system:

$\frac{{\partial u}}{{\partial y}} = \frac{\partial }{{\partial y}}\Big[ {x + \frac{2}{3}{{\left( {{x^2} - {y^2}} \right)}^{\frac{3}{2}}} + \varphi \left( y \right)} \Big] = - 2y\sqrt {{x^2} - {y^2}} ,\;\; \Rightarrow - \cancel{2y\sqrt {{x^2} - {y^2}}} + \varphi'\left( y \right) = - \cancel{2y\sqrt {{x^2} - {y^2}}} ,\;\; \Rightarrow \varphi'\left( y \right) = 0.$

By integrating the last expression, we find the function $$\varphi \left( y \right):$$

$\varphi \left( y \right) = C,$

where $$C$$ is a constant.

Thus, the general solution of the given differential equation has the form:

$x + \frac{2}{3}{\left( {{x^2} - {y^2}} \right)^{\frac{3}{2}}} + C = 0.$

### Example 6.

Solve the differential equation $\frac{1}{{{y^2}}} - \frac{2}{x} = \frac{{2xy'}}{{{y^3}}}$ with the initial condition $$y\left( 1 \right) = 1.$$

Solution.

Check the equation for exactness by converting it into standard form:

$\frac{1}{{{y^2}}} - \frac{2}{x} = \frac{{2x}}{{{y^3}}}\frac{{dy}}{{dx}},\;\; \Rightarrow \left( {\frac{1}{{{y^2}}} - \frac{2}{x}} \right)dx = \frac{{2x}}{{{y^3}}}dy,\;\; \Rightarrow \left( {\frac{1}{{{y^2}}} - \frac{2}{x}} \right)dx - \frac{{2x}}{{{y^3}}}dy = 0.$

The partial derivatives are

$\frac{{\partial Q}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( { - \frac{{2x}}{{{y^3}}}} \right) = - \frac{2}{{{y^3}}},\;\;\; \frac{{\partial P}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {\frac{1}{{{y^2}}} - \frac{2}{x}} \right) = - \frac{2}{{{y^3}}}.$

Hence, the given equation is exact. Therefore, we can write the following system of equations to determine the function $$u\left( {x,y} \right):$$

$\left\{ \begin{array}{l} \frac{{\partial u}}{{\partial x}} = \frac{1}{{{y^2}}} - \frac{2}{x}\\ \frac{{\partial u}}{{\partial y}} = - \frac{{2x}}{{{y^3}}} \end{array} \right..$

In the given case, it is more convenient to integrate the second equation with respect to the variable $$y:$$

$u\left( {x,y} \right) = \int {\left( { - \frac{{2x}}{{{y^3}}}} \right)dy} = \frac{x}{{{y^2}}} + \psi \left( x \right).$

Now we differentiate this expression with respect to the variable $$x:$$

$\frac{{\partial u}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {\frac{x}{{{y^2}}} + \psi \left( x \right)} \right] = \frac{1}{{{y^2}}} - \frac{2}{x},\;\; \Rightarrow \cancel{\frac{1}{{{y^2}}}} + \psi'\left( x \right) = \cancel{\frac{1}{{{y^2}}}} - \frac{2}{x},\;\; \Rightarrow \psi'\left( x \right) = - \frac{2}{x},\;\; \Rightarrow \psi \left( x \right) = - 2\ln \left| x \right| = \ln \frac{1}{{{x^2}}}.$

Thus, the general solution of the differential equation in implicit form is given by the expression:

$\frac{x}{{{y^2}}} + \ln \frac{1}{{{x^2}}} = C.$

The particular solution can be found using the initial condition $$y\left( 1 \right) = 1.$$ By substituting the initial values, we find the constant $$C:$$

$\frac{1}{{{1^2}}} + \ln \frac{1}{{{1^2}}} = C,\;\; \Rightarrow 1 + 0 = C,\;\; \Rightarrow C = 1.$

Hence, the solution of the given initial value problem is

$\frac{1}{{{y^2}}} + \ln \frac{1}{{{x^2}}} = 1.$