Riccati Equation
Solved Problems
Example 3.
Find the general solution of the differential equation \[{x^3}y' + {x^2}y - {y^2} = 2{x^4}.\]
Solution.
We convert this equation into the standard form:
\[{x^3}y' + {x^2}y - {y^2} = 2{x^4},\;\; \Rightarrow y' + \frac{y}{x} - \frac{{{y^2}}}{{{x^3}}} = 2x.\]
As you can see, we have a Riccati equation. Try to find a particular solution in the form \({y_1} = c{x^2}.\) Substituting this into the Riccati equation, we can determine the coefficient \(c:\)
\[{\left( {c{x^2}} \right)^\prime } + \frac{{c{x^2}}}{x} - \frac{{{{\left( {c{x^2}} \right)}^2}}}{{{x^3}}} = 2x,\;\; \Rightarrow 2cx + cx - {c^2}x = 2x,\;\; \Rightarrow 3c - {c^2} = 2,\;\; \Rightarrow {c^2} - 3c + 2 = 0.\]
Solving this quadratic equation, we obtain the value of \(c:\)
\[D = 9 - 4 \cdot 2 = 1,\;\; \Rightarrow {c_{1,2}} = \frac{{3 \pm \sqrt 1 }}{2} = 1,2.\]
Thus, there are even two particular solutions. However, we need only one of them. So we take, for example, \({y_1} = {x^2}.\)
As a result, we can write the general solution of the Riccati equation in the form:
\[y = {y_1} + u = {x^2} + u.\]
We get the following differential equation for the new function \(u\left( x \right):\)
\[{\left( {{x^2} + u} \right)^\prime } + \frac{{{x^2} + u}}{x} - \frac{{{{\left( {{x^2} + u} \right)}^2}}}{{{x^3}}} = 2x,\;\; \Rightarrow \cancel{2x} + u' + x + \frac{u}{x} - \frac{{{x^4} + 2u{x^2} + {u^2}}}{{{x^3}}} - \cancel{2x} = 0,\;\; \Rightarrow u' + \cancel{x} + \frac{u}{x} - \cancel{x} - \frac{{2u}}{x} - \frac{{{u^2}}}{{{x^3}}} = 0,\;\; \Rightarrow u' - \frac{u}{x} = \frac{{{u^2}}}{{{x^3}}},\]
which is a Bernoulli equation. The substitution of \(z = {u^{ - 1}} = \frac{1}{u}\) converts it to a linear differential equation:
\[z' = {\left( {\frac{1}{u}} \right)^\prime } = - \frac{{u'}}{{{u^2}}},\;\; \Rightarrow u' - \frac{u}{x} = \frac{{{u^2}}}{{{x^3}}},\;\; \Rightarrow \frac{{u'}}{{{u^2}}} - \frac{1}{{xu}} = \frac{1}{{{x^3}}},\;\; \Rightarrow - \frac{{u'}}{{{u^2}}} + \frac{1}{{xu}} = - \frac{1}{{{x^3}}},\;\; \Rightarrow z' + \frac{z}{x} = - \frac{1}{{{x^3}}}.\]
Then we calculate the integrating factor to solve the linear equation:
\[v\left( x \right) = {e^{\int {\frac{1}{x}dx} }} = {e^{\ln \left| x \right|}} = \left| x \right|.\]
One can take the function \(v\left( x \right) = x\) as the integrating factor. Indeed, we can check that the left side of the equation becomes equal to the derivative of the product \(z\left( x \right)v\left( x \right)\) after multiplying by \(v\left( x \right) = x:\)
\[z = \frac{{\int {v\left( x \right)f\left( x \right)dx} + C}}{{v\left( x \right)}} = \frac{{\int {x \cdot \left( { - \frac{1}{{{x^3}}}} \right)dx} + C}}{x} = \frac{{ - \int {\frac{{dx}}{{{x^2}}}} + C}}{x} = \frac{{\frac{1}{x} + C}}{x} = \frac{1}{{{x^2}}} + \frac{C}{x} = \frac{{Cx + 1}}{{{x^2}}}.\]
Since \(z = \frac{1}{u},\) the function \(u\left( x \right)\) is defined by the formula
\[u\left( x \right) = \frac{1}{z} = \frac{{{x^2}}}{{Cx + 1}}.\]
Hence, the general solution of the original Riccati equation is given by
\[y = {y_1} + u = {x^2} + \frac{{{x^2}}}{{Cx + 1}} = \frac{{{x^2}\left( {Cx + 1} \right) + {x^2}}}{{Cx + 1}} = \frac{{C{x^3} + 2{x^2}}}{{Cx + 1}},\]
where \(C\) is a constant.
Example 4.
Solve the equation \[y' + 6{y^2} = \frac{1}{{{x^2}}}.\]
Solution.
As it can be seen, this is a special Riccati equation of type \(y' = b{y^2} + c{x^n}\) with the degree \(n = -2.\)
By making the substitution \(y = \frac{1}{z}\) we can convert the equation to a homogeneous one and then integrate.
Let \(z = \frac{1}{y},\) \(z' = - \frac{{y'}}{{{y^2}}}.\) Then
\[y' + 6{y^2} = \frac{1}{{{x^2}}},\;\; \Rightarrow y' = - 6{y^2} + \frac{1}{{{x^2}}},\;\; \Rightarrow - \frac{{y'}}{{{y^2}}} = 6 - \frac{1}{{{y^2}{x^2}}},\;\; \Rightarrow z' = 6 - \frac{{{z^2}}}{{{x^2}}},\;\; \Rightarrow z' = 6 - {\left( {\frac{z}{x}} \right)^2}.\]
To solve the homogeneous equation we make one more substitution: \(z = tx,\;z' = t'x + t.\) Hence,
\[t'x + t = 6 - {t^2},\;\; \Rightarrow x\frac{{dt}}{{dx}} = 6 - t - {t^2},\;\; \Rightarrow \frac{{dt}}{{{t^2} + t - 6}} = - \frac{{dx}}{x},\;\; \Rightarrow \int {\frac{{dt}}{{{t^2} + t - 6}}} = - \int {\frac{{dx}}{x}} .\]
The trinomial in the denominator of the left side can be factored as follows:
\[{t^2} + t - 6 = \left( {t + 3} \right)\left( {t - 2} \right),\]
so we may use partial decomposition to simplify the integrand:
\[
\frac{1}{{{t^2} + t - 6}} = \frac{1}{{\left( {t + 3} \right)\left( {t - 2} \right)}} = \frac{A}{{t + 3}} + \frac{B}{{t - 2}},\;\; \Rightarrow
A\left( {t - 2} \right) + B\left( {t + 3} \right) = 1,\;\; \Rightarrow
At - 2A + Bt + 3B = 1,\;\; \Rightarrow
\left( {A + B} \right)t + 3B - 2A = 1,\;\; \Rightarrow
\left\{ {\begin{array}{*{20}{l}}
{A + B = 0}\\
{3B - 2A = 1}
\end{array}} \right.,\;\; \Rightarrow
\left\{ {\begin{array}{*{20}{l}}
{A = - B}\\
{5B = 1}
\end{array}} \right.,\;\; \Rightarrow
\left\{ {\begin{array}{*{20}{l}}
{A = - \frac{1}{5}}\\
{B = \frac{1}{5}}
\end{array}} \right..\]
As a result, we have:
\[\int {\frac{{dt}}{{\left( {t + 3} \right)\left( {t - 2} \right)}}} = - \int {\frac{{dx}}{x}} ,\;\; \Rightarrow - \frac{1}{5}\int {\frac{{dt}}{{t + 3}}} + \frac{1}{5}\int {\frac{{dt}}{{t - 2}}} = - \int {\frac{{dx}}{x}} ,\;\; \Rightarrow \frac{1}{5}\ln \left| {t + 3} \right| - \frac{1}{5}\ln \left| {t - 2} \right| = \ln \left| x \right| + \ln {C_1}\;\left( {{C_1} \gt 0} \right),\;\; \Rightarrow \frac{1}{5}\ln \left| {\frac{{t + 3}}{{t - 2}}} \right| = \ln \left( {{C_1}\left| x \right|} \right),\;\; \Rightarrow \ln \left| {\frac{{t + 3}}{{t - 2}}} \right| = \ln \left( {C_1^5{{\left| x \right|}^5}} \right),\;\; \Rightarrow \left| {\frac{{t + 3}}{{t - 2}}} \right| = C_1^5{\left| x \right|^5},\;\; \Rightarrow \frac{{t + 3}}{{t - 2}} = \pm C_1^5{x^5}.\]
Rename the constant: \(C = \pm C_1^5,\) so the solution for the function \(t\left( x \right)\) will have the form:
\[\frac{{t + 3}}{{t - 2}} = C{x^5}.\]
Remember that \(t = \frac{z}{x}.\) Therefore,
\[\frac{{\frac{z}{x} + 3}}{{\frac{z}{x} - 2}} = C{x^5},\;\; \Rightarrow \frac{{z + 3x}}{{z - 2x}} = C{x^5}.\]
Returning to the variable \(y,\) which is related to \(z\) by the relationship \(z = \frac{1}{y},\) we get
\[\frac{{\frac{1}{y} + 3x}}{{\frac{1}{y} - 2x}} = C{x^5},\;\; \Rightarrow \frac{{1 + 3xy}}{{1 - 2xy}} = C{x^5}.\]
The last expression is the general solution of the Riccati equation in the implicit form. Here the constant \(C\) is any real number. Indeed, substituting \(C = 0,\) we see that this value also satisfies the differential equation:
\[C = 0,\;\; \Rightarrow 1 + 3xy = 0,\;\; \Rightarrow y = - \frac{1}{{3x}},\;\; \Rightarrow y' = \frac{1}{{3{x^2}}}.\]
Hence,
\[\frac{1}{{3{x^2}}} + 6{\left( { - \frac{1}{{3x}}} \right)^2} = \frac{1}{{{x^2}}},\;\; \Rightarrow \frac{1}{{3{x^2}}} + 6 \cdot \frac{1}{{9{x^2}}} = \frac{1}{{{x^2}}},\;\; \Rightarrow \frac{1}{{3{x^2}}} + \frac{2}{{3{x^2}}} = \frac{1}{{{x^2}}},\;\; \Rightarrow \frac{1}{{{x^2}}} \equiv \frac{1}{{{x^2}}}.\]
