Internal problem ID [8357]
Internal file name [OUTPUT/7290_Sunday_June_05_2022_05_42_50_PM_31989286/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 20.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[_Riccati]
\[ \boxed {y^{\prime }-y^{2}+\left (x^{2}+1\right ) y=2 x} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -x^{2} y +y^{2}+2 x -y \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -x^{2} y +y^{2}+2 x -y \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=2 x\), \(f_1(x)=-x^{2}-1\) 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^{2}-1\\ f_2^2 f_0 &=2 x \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )-\left (-x^{2}-1\right ) u^{\prime }\left (x \right )+2 x u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \left (c_{1} \left (\int {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}d x \right )+c_{2} \right ) {\mathrm e}^{-\frac {x \left (x^{2}+3\right )}{3}} \] The above shows that \[ u^{\prime }\left (x \right ) = -{\mathrm e}^{-\frac {x \left (x^{2}+3\right )}{3}} c_{1} \left (x^{2}+1\right ) \left (\int {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}d x \right )+\left (-x^{2}-1\right ) c_{2} {\mathrm e}^{-\frac {x \left (x^{2}+3\right )}{3}}+c_{1} \] Using the above in (1) gives the solution \[ y = -\frac {\left (-{\mathrm e}^{-\frac {x \left (x^{2}+3\right )}{3}} c_{1} \left (x^{2}+1\right ) \left (\int {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}d x \right )+\left (-x^{2}-1\right ) c_{2} {\mathrm e}^{-\frac {x \left (x^{2}+3\right )}{3}}+c_{1} \right ) {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}}{c_{1} \left (\int {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}d x \right )+c_{2}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (\int {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}d x \right ) \left (x^{2}+1\right ) c_{1} +c_{2} x^{2}-c_{1} {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}+c_{2}}{c_{1} \left (\int {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}d x \right )+c_{2}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (\int {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}d x \right ) \left (x^{2}+1\right ) c_{1} +c_{2} x^{2}-c_{1} {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}+c_{2}}{c_{1} \left (\int {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}d x \right )+c_{2}} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (\int {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}d x \right ) \left (x^{2}+1\right ) c_{1} +c_{2} x^{2}-c_{1} {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}+c_{2}}{c_{1} \left (\int {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}d x \right )+c_{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}+\left (x^{2}+1\right ) y=2 x \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}-\left (x^{2}+1\right ) y+2 x \end {array} \]
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`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 67
dsolve(diff(y(x),x) - y(x)^2 +(x^2 + 1)*y(x) - 2*x=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {-x^{2} \left (\int {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}d x \right )+c_{1} x^{2}+{\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}-\left (\int {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}d x \right )+c_{1}}{c_{1} -\left (\int {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}d x \right )} \]
✓ Solution by Mathematica
Time used: 0.294 (sec). Leaf size: 58
DSolve[y'[x] - y[x]^2 +(x^2 + 1)*y[x] - 2*x==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {e^{\frac {x^3}{3}+x}}{-\int _1^xe^{\frac {K[1]^3}{3}+K[1]}dK[1]+c_1}+x^2+1 \\ y(x)\to x^2+1 \\ \end{align*}