First Integrals
Solved Problems
Example 1.
Solve the system of equations
\[\frac{{dx}}{{dt}} = \frac{1}{y},\;\; \frac{{dy}}{{dt}} = - \frac{1}{x}.\]
Solution.
We write the system in the form
\[\left\{ \begin{array}{l}
ydx = dt\\
xdy = - dt
\end{array} \right..\]
Adding the two equations, we obtain
\[ydx + xdy = 0,\;\; \Rightarrow d\left( {xy} \right) = 0.\]
Hence, we find the first integral:
\[xy = {C_1},\]
where \({C_1}\) is any number other than zero.
Express the solutions \(x\left( t \right),y\left( t \right)\) explicitly. Substitute the function \(y = {\frac{{{C_1}}}{x}}\) into the first equation and integrate it:
\[\frac{{dx}}{{dt}} = \frac{1}{y} = \frac{x}{{{C_1}}},\;\; \Rightarrow
\frac{{dx}}{x} = \frac{{dt}}{{{C_1}}},\;\; \Rightarrow
\ln \left| x \right| = \frac{t}{{{C_1}}} + \ln {C_2} = \ln {e^{\frac{t}{{{C_1}}}}} + \ln {C_2} = \ln \left( {{C_2}{e^{\frac{t}{{{C_1}}}}}} \right),\;\; \Rightarrow
x\left( t \right) = {C_2}{e^{\frac{t}{{{C_1}}}}},\]
where \({C_2} \ne 0\) is an arbitrary constant.
Now we find the solution \(y\left( t \right):\)
\[xy = {C_1},\;\; \Rightarrow
y\left( t \right) = \frac{{{C_1}}}{{x\left( t \right)}}
= \frac{{{C_1}}}{{{C_2}}}{e^{ - \frac{t}{{{C_1}}}}}.\]
The final answer is
\[x\left( t \right) = {C_2}{e^{\frac{t}{{{C_1}}}}},\;\;
y\left( t \right) = \frac{{{C_1}}}{{{C_2}}}{e^{ - \frac{t}{{{C_1}}}}},\;\;
{C_1} \ne 0,\;{C_2} \ne 0.\]
Example 2.
Solve the system of equations
\[
\frac{{dx}}{{dt}} = {x^2}y,\;
\frac{{dy}}{{dt}} = x{y^2},\;
x \gt 0,\;y \gt 0.\]
Solution.
We write the system in the symmetric form:
\[\frac{{dx}}{{{x^2}y}} = \frac{{dy}}{{x{y^2}}}.\]
Dividing both sides by \(\frac{1}{{xy}},\) we get an equation that can be integrated:
\[
\frac{{dx}}{x} = \frac{{dy}}{y},\;\; \Rightarrow
\ln \left| x \right| = \ln \left| y \right| + \ln {C_1},\;\; \Rightarrow
x = {C_1}y.\]
This relationship is a first integral of the system. We express the variable \(y\) in terms of \(x\) and substitute into the first equation of the system:
\[
y = \frac{x}{{{C_1}}},\;\; \Rightarrow \frac{{dx}}{{dt}} = {x^2}y = \frac{{{x^3}}}{{{C_1}}},\;\; \Rightarrow
\frac{{dx}}{{{x^3}}} = \frac{{dt}}{{{C_1}}},\;\; \Rightarrow
- \frac{1}{{2{x^2}}} = \frac{t}{{{C_1}}} + {C_2},\;\; \Rightarrow
2{x^2} = \frac{{{C_1}}}{{ - t - {C_1}{C_2}}},\;\; \Rightarrow
{x^2} = \frac{{{C_1}}}{{ - 2t - 2{C_1}{C_2}}}.\]
Replace \({ - 2{C_1}{C_2}}\) with \({C_2}:\)
\[{x^2} = \frac{{{C_1}}}{{{C_2} - 2t}},\;\; \Rightarrow
x = \pm \sqrt {\frac{{{C_1}}}{{{C_2} - 2t}}} .
\]
The correct solution is the positive root as \(x \gt 0\) by the problem condition:
\[x\left( t \right) = \sqrt {\frac{{{C_1}}}{{{C_2} - 2t}}} .\]
Now we find the function \(y\left( t \right):\)
\[
y\left( t \right) = \frac{{x\left( t \right)}}{{{C_1}}}
= \frac{1}{{{C_1}}}\sqrt {\frac{{{C_1}}}{{{C_2} - 2t}}}
= \frac{1}{{\sqrt {{C_1}\left( {{C_2} - 2t} \right)} }}.\]
So, the general solution is given by
\[
x\left( t \right) = \sqrt {\frac{{{C_1}}}{{{C_2} - 2t}}} ,\;\;
y\left( t \right) = \frac{1}{{\sqrt {{C_1}\left( {{C_2} - 2t} \right)} }}.\]
Example 3.
Find the general solution of the system of differential equations
\[
\frac{{dx}}{{dt}} = \frac{y}{z},\;
\frac{{dy}}{{dt}} = \frac{x}{z},\;
\frac{{dz}}{{dt}} = \frac{x}{y},\;
x \gt 0,\;y \gt 0,\;z \gt 0.
\]
Solution.
We transform the equations of the system to get integrable combinations. Divide the second equation by the first:
\[
\frac{{dy}}{{dt}}:\frac{{dx}}{{dt}} = \frac{x}{z}:\frac{y}{z},\;\; \Rightarrow
\frac{{dy}}{{dx}} = \frac{x}{y},\;\; \Rightarrow
ydy = xdx,\;\; \Rightarrow
{y^2} - {x^2} = {C_1}.\]
The result is a first integral.
Similarly, dividing the third equation by the second equation, we obtain one more first integral:
\[
\frac{{dz}}{{dt}}:\frac{{dy}}{{dt}} = \frac{x}{y}:\frac{x}{z},\;\; \Rightarrow
\frac{{dz}}{{dy}} = \frac{z}{y},\;\; \Rightarrow
\frac{{dz}}{z} = \frac{{dy}}{y},\;\; \Rightarrow
\ln \left| z \right| = \ln \left| y \right| + \ln {C_2},\;\; \Rightarrow
z = {C_2}y.\]
Obviously, these two first integrals are independent.
Now we find the solutions \(x\left( t \right),\) \(y\left( t \right),\) \(z\left( t \right)\) explicitly. Substitute into the first equation the expressions for \(x\) and \(z,\) respectively, from the first and second integral:
\[
{x^2} = {y^2} - {C_1},\;\; \Rightarrow
x = \sqrt {{y^2} - {C_1}} ,\;\; \Rightarrow
\frac{{dx}}{{dt}} = \frac{y}{{\sqrt {{y^2} - {C_1}} }}\frac{{dy}}{{dt}},\;\; \Rightarrow
\frac{y}{{\sqrt {{y^2} - {C_1}} }}\frac{{dy}}{{dt}} = \frac{\cancel{y}}{{{C_2}\cancel{y}}},\;\; \Rightarrow
\frac{{d\left( {{y^2} - {C_1}} \right)}}{{2\sqrt {{y^2} - {C_1}} }} = \frac{{dt}}{{{C_2}}},\;\; \Rightarrow
\int {\frac{{d\left( {{y^2} - {C_1}} \right)}}{{2\sqrt {{y^2} - {C_1}} }}} = \int {\frac{{dt}}{{{C_2}}}} ,\;\; \Rightarrow
\sqrt {{y^2} - {C_1}} = \frac{t}{{{C_2}}} + {C_3},\;\; \Rightarrow
x\left( t \right) = \frac{t}{{{C_2}}} + {C_3}.\]
Then it is easy to determine the solutions \(y\left( t \right)\) and \(z\left( t \right):\)
\[
y\left( t \right) = \sqrt {{{\left( {\frac{t}{{{C_2}}} + {C_3}} \right)}^2} + {C_1}}
= \frac{1}{{{C_2}}}\sqrt {{{\left( {t + {C_2}{C_3}} \right)}^2} + {C_1}C_2^2} ,\;\;
z\left( t \right) = {C_2}y\left( t \right)
= \sqrt {{{\left( {t + {C_2}{C_3}} \right)}^2} + {C_1}C_2^2} .\]
Example 4.
Find the general solution of the system using the first integrals.
\[
\frac{{dx}}{{dt}} = y + z,\;
\frac{{dy}}{{dt}} = x - z,\;
\frac{{dz}}{{dt}} = x + y.\]
Solution.
We write the system in the symmetric form:
\[\frac{{dx}}{{y + z}} = \frac{{dy}}{{x - z}} = \frac{{dz}}{{x + y}}.\]
Using the equal fractions property, we get
\[
\frac{{dx + dy}}{{y + \cancel{z} + x - \cancel{z}}} = \frac{{dz}}{{x + y}},\;\; \Rightarrow
\frac{{d\left( {x + y} \right)}}{{x + y}} = \frac{{dz}}{{x + y}},\;\; \Rightarrow
d\left( {x + y} \right) = dz,\;\; \Rightarrow
x + y = z + {C_1}\;\; \text{or}\;\; x + y - z = {C_1}.\]
Another integrable combination can be obtained through the following transformations:
\[
\frac{{xdx - ydy}}{{x\left( {y + z} \right) - y\left( {x - z} \right)}} = \frac{{dz}}{{x + y}},\;\; \Rightarrow
\frac{{\frac{1}{2}d\left( {{x^2} - {y^2}} \right)}}{{\cancel{xy} + xz - \cancel{yx} + yz}} = \frac{{dz}}{{x + y}},\;\; \Rightarrow
\frac{{\frac{1}{2}d\left( {{x^2} - {y^2}} \right)}}{{z\left( {x + y} \right)}} = \frac{{dz}}{{x + y}},\;\; \Rightarrow
\frac{1}{2}d\left( {{x^2} - {y^2}} \right) = zdz,\;\; \Rightarrow
d\left( {{x^2} - {y^2}} \right) = d{z^2},\;\; \Rightarrow
{x^2} - {y^2} = {z^2} + {C_2}\;\; \Rightarrow
\text{or}\;\;{x^2} - {y^2} - {z^2} = {C_2}.\]
So, we have found two first integrals:
\[
{U_1} = x + y - z = {C_1},\;\;
{U_2} = {x^2} - {y^2} - {z^2} = {C_2}.\]
Let us verify that these integrals are independent. Compute the rank of the Jacobian matrix:
\[
\text{rank}\left( J \right)
= \text{rank}\left[ {\begin{array}{*{20}{c}}
{\frac{{\partial {U_1}}}{{\partial x}}}&{\frac{{\partial {U_1}}}{{\partial y}}}&{\frac{{\partial {U_1}}}{{\partial z}}}\\
{\frac{{\partial {U_2}}}{{\partial x}}}&{\frac{{\partial {U_2}}}{{\partial y}}}&{\frac{{\partial {U_2}}}{{\partial z}}}
\end{array}} \right]
= \text{rank}\left[ {\begin{array}{*{20}{c}}
1&1&{ - 1}\\
{2x}&{ - 2y}&{ - 2z}
\end{array}} \right] = 2,\]
that is the rank is equal to the number of first integrals. Therefore, the above integrals define the general solution in implicit form.
