Differential Equations

First Order Equations

1st Order Diff Equations Logo

Using an Integrating Factor

Solved Problems

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Example 3

Solve the differential equation

\[\left( {x{y^2} - 2{y^3}} \right)dx + \left( {3 - 2x{y^2}} \right)dy = 0.\]

Example 4

Solve the equation \[\left( {xy + 1} \right)dx + {x^2}dy = 0.\]

Example 5

Solve the equation \[ydx + \left( {{x^2} + {y^2} - x} \right)dy = 0\] using the integrating factor \(\mu \left( {x,y} \right) = {x^2} + {y^2}.\)

Example 3.

Solve the differential equation

\[\left( {x{y^2} - 2{y^3}} \right)dx + \left( {3 - 2x{y^2}} \right)dy = 0.\]

Solution.

The given equation is not exact, because

\[\frac{{\partial Q}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {3 - 2x{y^2}} \right) = - 2{y^2} \ne \frac{{\partial P}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {x{y^2} - 2{y^3}} \right) = 2xy - 6{y^2}.\]

We try to find the general solution of the equation using an integrating factor. Calculate the difference

\[\frac{{\partial P}}{{\partial y}} - \frac{{\partial Q}}{{\partial x}} = 2xy - 6{y^2} - \left( { - 2{y^2}} \right) = 2xy - 4{y^2}.\]

Notice that the expression

\[\frac{1}{P}\left( {\frac{{\partial P}}{{\partial y}} - \frac{{\partial Q}}{{\partial x}}} \right) = \frac{{2xy - 4{y^2}}}{{x{y^2} - 2{y^3}}} = \frac{{2\cancel{\left( {xy - 2{y^2}} \right)}}}{{y\cancel{\left( {xy - 2{y^2}} \right)}}} = \frac{2}{y} \]

depends only on \(y.\) Therefore, the integrating factor \(\mu\) is also a function only of the variable \(y.\) We can find it from the equation

\[\frac{1}{\mu }\frac{{d\mu }}{{dy}} = - \frac{1}{P}\left( {\frac{{\partial P}}{{\partial y}} - \frac{{\partial Q}}{{\partial x}}} \right) = - \frac{2}{y}.\]

Integrating, we get:

\[\int {\frac{{d\mu }}{\mu }} = - 2\int {\frac{{dy}}{y}} ,\;\; \Rightarrow \ln \left| \mu \right| = - 2\ln \left| \mu \right|,\;\; \Rightarrow \mu = \pm \frac{1}{{{y^2}}}.\]

By choosing \(\mu = {\frac{1}{{{y^2}}}}\) as the integrating factor and then multiplying the original differential equation by it, we get the exact equation:

\[\left( {x - 2y} \right)dx + \left( {\frac{3}{{{y^2}}} - 2x} \right)dy = 0.\]

Indeed, we see that

\[\frac{{\partial Q}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {\frac{3}{{{y^2}}} - 2x} \right) = - 2 = \frac{{\partial P}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {x - 2y} \right) = - 2.\]

Notice that when we multiplied by the integrating factor we lost the solution \(y = 0.\) This can be proved by direct substitution of the solution \(y = 0\) in the original differential equation.

Now we find the function \(u\) from the system of equations:

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

It follows from the first equation that

\[u\left( {x,y} \right) = \int {\left( {x - 2y} \right)dx} = \frac{{{x^2}}}{2} - 2yx + \varphi \left( y \right).\]

Then we get from the second equation:

\[\frac{{\partial u}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {\frac{{{x^2}}}{2} - 2yx + \varphi \left( y \right)} \right] = \frac{3}{{{y^2}}} - 2x,\;\; \Rightarrow - 2x + \varphi'\left( y \right) = \frac{3}{{{y^2}}} - 2x,\;\; \Rightarrow \varphi'\left( y \right) = \frac{3}{{{y^2}}},\;\; \Rightarrow \varphi \left( y \right) = \int {\frac{3}{{{y^2}}}dy} = - \frac{3}{y}.\]

Thus, the original differential equation has the following solutions:

\[\frac{{{x^2}}}{2} - 2yx - \frac{3}{y} = C,\;\; y = 0,\]

where \(C\) is a constant.

Example 4.

Solve the equation \[\left( {xy + 1} \right)dx + {x^2}dy = 0.\]

Solution.

First of all we check this equation for exactness:

\[\frac{{\partial Q}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {{x^2}} \right) = 2x\; \ne \frac{{\partial P}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {xy + 1} \right) = x.\]

The partial derivatives are not equal to each other. Therefore, this equation is not exact. Calculate the difference of the derivatives:

\[\frac{{\partial P}}{{\partial y}} - \frac{{\partial Q}}{{\partial x}} = x - 2x = - x.\]

Now we try to use the integrating factor in the form \(z = xy.\) Here we have

\[\frac{{\partial z}}{{\partial x}} = y,\;\; \frac{{\partial z}}{{\partial y}} = x.\]

Then

\[Q\frac{{\partial z}}{{\partial x}} - P\frac{{\partial z}}{{\partial y}} = {x^2} \cdot y - \left( {xy + 1} \right) \cdot x = \cancel{{x^2}y} - \cancel{{x^2}y} - x = - x,\]

and hence we get

\[\frac{1}{\mu }\frac{{d\mu }}{{dz}} = \frac{{\frac{{\partial P}}{{\partial y}} - \frac{{\partial Q}}{{\partial x}}}}{{Q\frac{{\partial z}}{{\partial x}} - P\frac{{\partial z}}{{\partial y}}}} = \frac{{ - x}}{{ - x}} = 1.\]

We see that the integrating factor depends only on \(z:\)

\[\mu \left( {x,y} \right) = \mu \left( z \right) = \mu \left( {xy} \right).\]

We can find it by integrating the last equation:

\[\frac{1}{\mu }\frac{{d\mu }}{{dz}} = 1,\;\; \Rightarrow \int {\frac{{d\mu }}{\mu }} = \int {dz} ,\;\; \Rightarrow \ln \left| \mu \right| = z,\;\; \Rightarrow \mu = {e^{ \pm z}} = {e^{ \pm xy}}.\]

By choosing the function \(\mu = {e^{xy}}\) we can convert the original differential equation into exact:

\[\left( {xy + 1} \right){e^{xy}}dx + {x^2}{e^{xy}}dy = 0.\]

Check this using again the test for exactness:

\[\frac{{\partial Q}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {{x^2}{e^{xy}}} \right] = 2x{e^{xy}} + {x^2}y{e^{xy}},\]
\[\frac{{\partial P}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {\left( {xy + 1} \right){e^{xy}}} \right] = x{e^{xy}} + \left( {xy + 1} \right)x{e^{xy}} = x{e^{xy}} + {x^2}y{e^{xy}} + x{e^{xy}} = 2x{e^{xy}} + {x^2}y{e^{xy}}.\]

As one can see, now this equation is exact. We find its general solution form the system of equations:

\[\left\{ \begin{array}{l} \frac{{\partial u}}{{\partial x}} = \left( {xy + 1} \right){e^{xy}}\\ \frac{{\partial u}}{{\partial y}} = {x^2}{e^{xy}} \end{array} \right..\]

Integrate the second equation with respect to the variable \(y\) (considering \(x\) as a constant):

\[u\left( {x,y} \right) = \int {{x^2}{e^{xy}}dy} = {x^2}\int {{e^{xy}}dy} = {x^2} \cdot \frac{1}{x}{e^{xy}} + \psi \left( x \right) = x{e^{xy}} + \psi \left( x \right).\]

Substitute this in the first equation of the system to get:

\[\frac{{\partial u}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {x{e^{xy}} + \psi \left( x \right)} \right] = \left( {xy + 1} \right){e^{xy}},\;\; \Rightarrow 1 \cdot {e^{xy}} + xy{e^{xy}} + \psi'\left( x \right) = \left( {xy + 1} \right){e^{xy}},\;\; \Rightarrow \left( {xy + 1} \right){e^{xy}} + \psi'\left( x \right) = \left( {xy + 1} \right){e^{xy}},\;\; \Rightarrow \psi'\left( x \right) = 0,\;\; \Rightarrow \psi \left( x \right) = C.\]

Hence, the general solution of the given differential equation is written in the form:

\[x{e^{xy}} + C = 0,\]

where \(C\) is any real number.

Example 5.

Solve the equation \[ydx + \left( {{x^2} + {y^2} - x} \right)dy = 0\] using the integrating factor \(\mu \left( {x,y} \right) = {x^2} + {y^2}.\)

Solution.

We can make sure that this equation is not exact:

\[\frac{{\partial Q}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {{x^2} + {y^2} - x} \right) = 2x - 1 \ne \frac{{\partial P}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( y \right) = 1.\]

The difference of the partial derivatives is

\[\frac{{\partial P}}{{\partial y}} - \frac{{\partial Q}}{{\partial x}} = 1 - \left( {2x - 1} \right) = 2 - 2x.\]

Using the integrating factor \(\mu = z = {x^2} + {y^2},\) we find that

\[\frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {{x^2} + {y^2}} \right) = 2x,\;\;\; \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {{x^2} + {y^2}} \right) = 2y.\]

Calculate the following expression:

\[Q\frac{{\partial z}}{{\partial x}} - P\frac{{\partial z}}{{\partial y}} = \left( {{x^2} + {y^2} - x} \right) \cdot 2x - y \cdot 2y = 2{x^3} + 2x{y^2} - 2{x^2} - 2{y^2} = 2x\left( {{x^2} + {y^2}} \right) - 2\left( {{x^2} + {y^2}} \right) = \left( {{x^2} + {y^2}} \right)\left( {2x - 2} \right).\]

As a result, we obtain the differential equation for the function \(\mu \left( z \right):\)

\[\frac{1}{\mu }\frac{{d\mu }}{{dz}} = \frac{{\frac{{\partial P}}{{\partial y}} - \frac{{\partial Q}}{{\partial x}}}}{{Q\frac{{\partial z}}{{\partial x}} - P\frac{{\partial z}}{{\partial y}}}} = \frac{{2 - 2x}}{{\left( {{x^2} + {y^2}} \right)\left( {2x - 2} \right)}} = - \frac{\cancel{2x - 2}}{{z \cancel{\left( {2x - 2} \right)}}} = - \frac{1}{z}.\]

By integrating we get the function \(\mu \left( z \right):\)

\[\int {\frac{{d\mu }}{\mu }} = - \int {\frac{{dz}}{z}} ,\;\; \Rightarrow \ln \left| \mu \right| = - \ln \left| z \right|,\;\; \Rightarrow \mu = \pm \frac{1}{z}.\]

We can choose the integrating factor \(\mu = \frac{1}{z} = \frac{1}{{{x^2} + {y^2}}}.\) After multiplication by \(\frac{1}{{{x^2} + {y^2}}}\) the original differential equation is converted into exact:

\[\frac{y}{{{x^2} + {y^2}}}dx + \frac{{{x^2} + {y^2} - x}}{{{x^2} + {y^2}}}dy = 0\;\; \text{or}\;\;\frac{y}{{{x^2} + {y^2}}}dx + \left( {1 - \frac{x}{{{x^2} + {y^2}}}} \right)dy = 0.\]

The general solution \(u\left( {x,y} \right) = C\) is defined by the following system of equations:

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

Integrating the first equation with respect to \(x\) produces:

\[u\left( {x,y} \right) = \int {\frac{y}{{{x^2} + {y^2}}}dx} = y\int {\frac{{dx}}{{{x^2} + {y^2}}}} = y \cdot \frac{1}{y}\arctan \frac{x}{y} + \varphi \left( y \right) = \arctan \frac{x}{y} + \varphi \left( y \right).\]

Substituting this in the second equation, we have:

\[\frac{{\partial u}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {\arctan \frac{x}{y} + \varphi \left( y \right)} \right] = 1 - \frac{x}{{{x^2} + {y^2}}},\;\; \Rightarrow \frac{1}{{1 + {{\left( {\frac{x}{y}} \right)}^2}}} \cdot \left( { - \frac{x}{{{y^2}}}} \right) + \varphi'\left( y \right) = 1 - \frac{x}{{{x^2} + {y^2}}},\;\; \Rightarrow - \frac{{\cancel{y^2}x}}{{\left( {{x^2} + {y^2}} \right)\cancel{y^2}}} + \varphi'\left( y \right) = 1 - \frac{x}{{{x^2} + {y^2}}},\;\; \Rightarrow \varphi'\left( y \right) = 1,\;\; \Rightarrow \varphi \left( y \right) = y.\]

Thus, the general solution of differential equation in implicit form is defined by the formula:

\[\arctan \frac{x}{y} + y = C,\]

where \(C\) is a constant.

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