Internal problem ID [3213]
Internal file name [OUTPUT/2705_Sunday_June_05_2022_08_39_01_AM_70722247/index.tex]
Book: Differential equations for engineers by Wei-Chau XIE, Cambridge Press 2010
Section: Chapter 2. First-Order and Simple Higher-Order Differential Equations. Page
78
Problem number: 71.1.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program :
Maple gives the following as the ode type
[_Riccati]
\[ \boxed {y^{\prime }+y^{2}=x^{2}+1} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= x^{2}-y^{2}+1 \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = x^{2}-y^{2}+1 \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=x^{2}+1\), \(f_1(x)=0\) 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 &=0\\ f_2^2 f_0 &=x^{2}+1 \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 \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{\frac {x^{2}}{2}} \left (c_{1} +\operatorname {erf}\left (x \right ) c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left (x \sqrt {\pi }\, \left (c_{1} +\operatorname {erf}\left (x \right ) c_{2} \right ) {\mathrm e}^{x^{2}}+2 c_{2} \right ) {\mathrm e}^{-\frac {x^{2}}{2}}}{\sqrt {\pi }} \] Using the above in (1) gives the solution \[ y = \frac {\left (x \sqrt {\pi }\, \left (c_{1} +\operatorname {erf}\left (x \right ) c_{2} \right ) {\mathrm e}^{x^{2}}+2 c_{2} \right ) \left ({\mathrm e}^{-\frac {x^{2}}{2}}\right )^{2}}{\sqrt {\pi }\, \left (c_{1} +\operatorname {erf}\left (x \right ) c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {2 \,{\mathrm e}^{-x^{2}}+x \sqrt {\pi }\, \left (c_{3} +\operatorname {erf}\left (x \right )\right )}{\sqrt {\pi }\, \left (c_{3} +\operatorname {erf}\left (x \right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {2 \,{\mathrm e}^{-x^{2}}+x \sqrt {\pi }\, \left (c_{3} +\operatorname {erf}\left (x \right )\right )}{\sqrt {\pi }\, \left (c_{3} +\operatorname {erf}\left (x \right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {2 \,{\mathrm e}^{-x^{2}}+x \sqrt {\pi }\, \left (c_{3} +\operatorname {erf}\left (x \right )\right )}{\sqrt {\pi }\, \left (c_{3} +\operatorname {erf}\left (x \right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }+y^{2}=x^{2}+1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-y^{2}+x^{2}+1 \end {array} \]
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 Special trying Riccati sub-methods: <- Riccati particular polynomial solution successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 37
dsolve(diff(y(x),x)+y(x)^2=1+x^2,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\sqrt {\pi }\, \operatorname {erf}\left (x \right ) x -2 c_{1} x +2 \,{\mathrm e}^{-x^{2}}}{\sqrt {\pi }\, \operatorname {erf}\left (x \right )-2 c_{1}} \]
✓ Solution by Mathematica
Time used: 0.136 (sec). Leaf size: 36
DSolve[y'[x]+y[x]^2==1+x^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to x+\frac {2 e^{-x^2}}{\sqrt {\pi } \text {erf}(x)+2 c_1} \\ y(x)\to x \\ \end{align*}