2.16 problem 41

Internal problem ID [3955]
Book : Ordinary differential equations and their solutions. By George Moseley Murphy. 1960
Section : Various 2
Problem number : 41
Date solved : Thursday, October 17, 2024 at 11:46:33 AM
CAS classification : [_Riccati]

Solve

\begin{align*} y^{\prime }+1-x&=\left (x +y\right ) y \end{align*}

2.16.1 Solved as first order ode of type Riccati

Time used: 0.405 (sec)

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= y x +y^{2}+x -1 \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = y x +y^{2}+x -1 \]

With Riccati ODE standard form

\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]

Shows that \(f_0(x)=x -1\), \(f_1(x)=x\) and \(f_2(x)=1\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u} \tag {1} \end{align*}

Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for \(u(x)\) which is

\begin{align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end{align*}

But

\begin{align*} f_2' &=0\\ f_1 f_2 &=x\\ f_2^2 f_0 &=x -1 \end{align*}

Substituting the above terms back in equation (2) gives

\begin{align*} u^{\prime \prime }\left (x \right )-x u^{\prime }\left (x \right )+\left (x -1\right ) u \left (x \right ) = 0 \end{align*}

Writing the ode as

\begin{align*} \frac {d^{2}u}{d x^{2}}-x \left (\frac {d u}{d x}\right )+\left (x -1\right ) u &= 0 \tag {1} \\ A \frac {d^{2}u}{d x^{2}} + B \frac {d u}{d x} + C u &= 0 \tag {2} \end{align*}

Comparing (1) and (2) shows that

\begin{align*} A &= 1 \\ B &= -x\tag {3} \\ C &= x -1 \end{align*}

Applying the Liouville transformation on the dependent variable gives

\begin{align*} z(x) &= u e^{\int \frac {B}{2 A} \,dx} \end{align*}

Then (2) becomes

\begin{align*} z''(x) = r z(x)\tag {4} \end{align*}

Where \(r\) is given by

\begin{align*} r &= \frac {s}{t}\tag {5} \\ &= \frac {2 A B' - 2 B A' + B^2 - 4 A C}{4 A^2} \end{align*}

Substituting the values of \(A,B,C\) from (3) in the above and simplifying gives

\begin{align*} r &= \frac {x^{2}-4 x +2}{4}\tag {6} \end{align*}

Comparing the above to (5) shows that

\begin{align*} s &= x^{2}-4 x +2\\ t &= 4 \end{align*}

Therefore eq. (4) becomes

\begin{align*} z''(x) &= \left ( \frac {1}{4} x^{2}-x +\frac {1}{2}\right ) z(x)\tag {7} \end{align*}

Equation (7) is now solved. After finding \(z(x)\) then \(u\) is found using the inverse transformation

\begin{align*} u &= z \left (x \right ) e^{-\int \frac {B}{2 A} \,dx} \end{align*}

The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases depending on the order of poles of \(r\) and the order of \(r\) at \(\infty \). The following table summarizes these cases.

Case	Allowed pole order for \(r\)	Allowed value for \(\mathcal {O}(\infty )\)
1	\(\left \{ 0,1,2,4,6,8,\cdots \right \} \)	\(\left \{ \cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \right \} \)
2	Need to have at least one pole that is either order \(2\) or odd order greater than \(2\). Any other pole order is allowed as long as the above condition is satisfied. Hence the following set of pole orders are all allowed. \(\{1,2\}\),\(\{1,3\}\),\(\{2\}\),\(\{3\}\),\(\{3,4\}\),\(\{1,2,5\}\).	no condition
3	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)

Table 1: Necessary conditions for each Kovacic case

The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\). Therefore

\begin{align*} O\left (\infty \right ) &= \text {deg}(t) - \text {deg}(s) \\ &= 0 - 2 \\ &= -2 \end{align*}

There are no poles in \(r\). Therefore the set of poles \(\Gamma \) is empty. Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(-2\) then the necessary conditions for case one are met. Therefore

\begin{align*} L &= [1] \end{align*}

Attempting to find a solution using case \(n=1\).

Since the order of \(r\) at \(\infty \) is \(O_r(\infty ) = -2\) then

\begin{alignat*}{3} v &= \frac {-O_r(\infty )}{2} &&= \frac {2}{2} &&= 1 \end{alignat*}

\([\sqrt r]_\infty \) is the sum of terms involving \(x^i\) for \(0\leq i \leq v\) in the Laurent series for \(\sqrt r\) at \(\infty \). Therefore

\begin{align*} [\sqrt r]_\infty &= \sum _{i=0}^{v} a_i x^i \\ &= \sum _{i=0}^{1} a_i x^i \tag {8} \end{align*}

Let \(a\) be the coefficient of \(x^v=x^1\) in the above sum. The Laurent series of \(\sqrt r\) at \(\infty \) is

\[ \sqrt r \approx \frac {x}{2}-1-\frac {1}{2 x}-\frac {1}{x^{2}}-\frac {9}{4 x^{3}}-\frac {11}{2 x^{4}}-\frac {57}{4 x^{5}}-\frac {77}{2 x^{6}} + \dots \tag {9} \]

Comparing Eq. (9) with Eq. (8) shows that

\[ a = {\frac {1}{2}} \]

From Eq. (9) the sum up to \(v=1\) gives

\begin{align*} [\sqrt r]_\infty &= \sum _{i=0}^{1} a_i x^i \\ &= \frac {x}{2}-1 \tag {10} \end{align*}

Now we need to find \(b\), where \(b\) be the coefficient of \(x^{v-1} = x^{0}=1\) in \(r\) minus the coefficient of same term but in \(\left ( [\sqrt r]_\infty \right )^2 \) where \([\sqrt r]_\infty \) was found above in Eq (10). Hence

\[ \left ( [\sqrt r]_\infty \right )^2 = \frac {1}{4} x^{2}-x +1 \]

This shows that the coefficient of \(1\) in the above is \(1\). Now we need to find the coefficient of \(1\) in \(r\). How this is done depends on if \(v=0\) or not. Since \(v=1\) which is not zero, then starting \(r=\frac {s}{t}\), we do long division and write this in the form

\[ r = Q + \frac {R}{t} \]

Where \(Q\) is the quotient and \(R\) is the remainder. Then the coefficient of \(1\) in \(r\) will be the coefficient this term in the quotient. Doing long division gives

\begin{align*} r &= \frac {s}{t} \\ &= \frac {x^{2}-4 x +2}{4} \\ &= Q + \frac {R}{4} \\ &= \left (\frac {1}{4} x^{2}-x +\frac {1}{2}\right ) + \left ( 0\right ) \\ &= \frac {1}{4} x^{2}-x +\frac {1}{2} \end{align*}

We see that the coefficient of the term \(\frac {1}{x}\) in the quotient is \(\frac {1}{2}\). Now \(b\) can be found.

\begin{align*} b &= \left ({\frac {1}{2}}\right )-\left (1\right )\\ &= -{\frac {1}{2}} \end{align*}

Hence

\begin{alignat*}{3} [\sqrt r]_\infty &= \frac {x}{2}-1\\ \alpha _{\infty }^{+} &= \frac {1}{2} \left ( \frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( \frac {-{\frac {1}{2}}}{{\frac {1}{2}}} - 1 \right ) &&= -1\\ \alpha _{\infty }^{-} &= \frac {1}{2} \left ( -\frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( -\frac {-{\frac {1}{2}}}{{\frac {1}{2}}} - 1 \right ) &&= 0 \end{alignat*}

The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\) is

\[ r=\frac {1}{4} x^{2}-x +\frac {1}{2} \]

Order of \(r\) at \(\infty \)	\([\sqrt r]_\infty \)	\(\alpha _\infty ^{+}\)	\(\alpha _\infty ^{-}\)
\(-2\)	\(\frac {x}{2}-1\)	\(-1\)	\(0\)

Now that the all \([\sqrt r]_c\) and its associated \(\alpha _c^{\pm }\) have been determined for all the poles in the set \(\Gamma \) and \([\sqrt r]_\infty \) and its associated \(\alpha _\infty ^{\pm }\) have also been found, the next step is to determine possible non negative integer \(d\) from these using

\begin{align*} d &= \alpha _\infty ^{s(\infty )} - \sum _{c \in \Gamma } \alpha _c^{s(c)} \end{align*}

Where \(s(c)\) is either \(+\) or \(-\) and \(s(\infty )\) is the sign of \(\alpha _\infty ^{\pm }\). This is done by trial over all set of families \(s=(s(c))_{c \in \Gamma \cup {\infty }}\) until such \(d\) is found to work in finding candidate \(\omega \). Trying \(\alpha _\infty ^{-} = 0\), and since there are no poles then

\begin{align*} d &= \alpha _\infty ^{-} \\ &= 0 \end{align*}

Since \(d\) an integer and \(d \geq 0\) then it can be used to find \(\omega \) using

\begin{align*} \omega &= \sum _{c \in \Gamma } \left ( s(c) [\sqrt r]_c + \frac {\alpha _c^{s(c)}}{x-c} \right ) + s(\infty ) [\sqrt r]_\infty \end{align*}

The above gives

\begin{align*} \omega &= (-) [\sqrt r]_\infty \\ &= 0 + (-) \left ( \frac {x}{2}-1 \right ) \\ &= 1-\frac {x}{2}\\ &= 1-\frac {x}{2} \end{align*}

Now that \(\omega \) is determined, the next step is find a corresponding minimal polynomial \(p(x)\) of degree \(d=0\) to solve the ode. The polynomial \(p(x)\) needs to satisfy the equation

\begin{align*} p'' + 2 \omega p' + \left ( \omega ' +\omega ^2 -r\right ) p = 0 \tag {1A} \end{align*}

Let

\begin{align*} p(x) &= 1\tag {2A} \end{align*}

Substituting the above in eq. (1A) gives

\begin{align*} \left (0\right ) + 2 \left (1-\frac {x}{2}\right ) \left (0\right ) + \left ( \left (-{\frac {1}{2}}\right ) + \left (1-\frac {x}{2}\right )^2 - \left (\frac {1}{4} x^{2}-x +\frac {1}{2}\right ) \right ) &= 0\\ 0 = 0 \end{align*}

The equation is satisfied since both sides are zero. Therefore the first solution to the ode \(z'' = r z\) is

\begin{align*} z_1(x) &= p e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int \left (1-\frac {x}{2}\right )d x}\\ &= {\mathrm e}^{-\frac {x \left (x -4\right )}{4}} \end{align*}

The first solution to the original ode in \(u\) is found from

\begin{align*} u_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dx} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {-x}{1} \,dx} \\ &= z_1 e^{\frac {x^{2}}{4}} \\ &= z_1 \left ({\mathrm e}^{\frac {x^{2}}{4}}\right ) \\ \end{align*}

Which simplifies to

\[ u_1 = {\mathrm e}^{x} \]

The second solution \(u_2\) to the original ode is found using reduction of order

\[ u_2 = u_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{u_1^2} \,dx \]

Substituting gives

\begin{align*} u_2 &= u_1 \int \frac { e^{\int -\frac {-x}{1} \,dx}}{\left (u_1\right )^2} \,dx \\ &= u_1 \int \frac { e^{\frac {x^{2}}{2}}}{\left (u_1\right )^2} \,dx \\ &= u_1 \left (-\frac {i \sqrt {\pi }\, {\mathrm e}^{-2} \sqrt {2}\, \operatorname {erf}\left (\frac {i \sqrt {2}\, x}{2}-i \sqrt {2}\right )}{2}\right ) \\ \end{align*}

Therefore the solution is

\begin{align*} u &= c_1 u_1 + c_2 u_2 \\ &= c_1 \left ({\mathrm e}^{x}\right ) + c_2 \left ({\mathrm e}^{x}\left (-\frac {i \sqrt {\pi }\, {\mathrm e}^{-2} \sqrt {2}\, \operatorname {erf}\left (\frac {i \sqrt {2}\, x}{2}-i \sqrt {2}\right )}{2}\right )\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

Taking derivative gives

\[ u^{\prime }\left (x \right ) = c_1 \,{\mathrm e}^{x}-\frac {i c_2 \sqrt {2}\, {\mathrm e}^{x -2} \sqrt {\pi }\, \operatorname {erf}\left (\frac {i \sqrt {2}\, \left (x -2\right )}{2}\right )}{2}+c_2 \,{\mathrm e}^{x -2} {\mathrm e}^{\frac {\left (x -2\right )^{2}}{2}} \]

Doing change of constants, the solution becomes

\[ y = -\frac {c_3 \,{\mathrm e}^{x}-\frac {i \sqrt {2}\, {\mathrm e}^{x -2} \sqrt {\pi }\, \operatorname {erf}\left (\frac {i \sqrt {2}\, \left (x -2\right )}{2}\right )}{2}+{\mathrm e}^{x -2} {\mathrm e}^{\frac {\left (x -2\right )^{2}}{2}}}{c_3 \,{\mathrm e}^{x}-\frac {i \sqrt {2}\, {\mathrm e}^{x -2} \sqrt {\pi }\, \operatorname {erf}\left (\frac {i \sqrt {2}\, \left (x -2\right )}{2}\right )}{2}} \]

Figure 69: Slope field plot
\(y^{\prime }+1-x = \left (x +y\right ) y\)

2.16.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )+1-x =y \left (x \right ) \left (x +y \left (x \right )\right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-1+x +y \left (x \right ) \left (x +y \left (x \right )\right ) \end {array} \]

2.16.3 Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
trying inverse linear 
trying homogeneous types: 
trying Chini 
differential order: 1; looking for linear symmetries 
trying exact 
Looking for potential symmetries 
trying Riccati 
trying Riccati sub-methods: 
   <- Riccati particular polynomial solution successful`
 

2.16.4 Maple dsolve solution

Solving time : 0.002 (sec)
Leaf size : 62

dsolve(diff(y(x),x)+1-x = y(x)*(x+y(x)), 
       y(x),singsol=all)
 

\[ y \left (x \right ) = \frac {-i \sqrt {\pi }\, {\mathrm e}^{-2} \sqrt {2}\, \operatorname {erf}\left (\frac {i \sqrt {2}\, \left (x -2\right )}{2}\right )+2 \,{\mathrm e}^{\frac {x \left (x -4\right )}{2}}-2 c_{1}}{i \sqrt {\pi }\, {\mathrm e}^{-2} \sqrt {2}\, \operatorname {erf}\left (\frac {i \sqrt {2}\, \left (x -2\right )}{2}\right )+2 c_{1}} \]

2.16.5 Mathematica DSolve solution

Solving time : 0.169 (sec)
Leaf size : 54

DSolve[{D[y[x],x]+1-x==(x+y[x])y[x],{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)\to -1+\frac {2 e^{\frac {1}{2} (x-2)^2}}{-\sqrt {2 \pi } \text {erfi}\left (\frac {x-2}{\sqrt {2}}\right )+2 e^2 c_1} \\ y(x)\to -1 \\ \end{align*}