Internal problem ID [6134]
Internal file name [OUTPUT/5382_Sunday_June_05_2022_03_35_50_PM_41186660/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 1. What is a differential equation. Section 1.2 THE NATURE OF SOLUTIONS.
Page 9
Problem number: 3(f).
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "quadrature"
Maple gives the following as the ode type
[_quadrature]
\[ \boxed {\left (1+x \right ) \left (x^{2}+1\right ) y^{\prime }=2 x^{2}+x} \] With initial conditions \begin {align*} [y \left (0\right ) = 1] \end {align*}
This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=0\\ q(x) &=\frac {x \left (2 x +1\right )}{\left (1+x \right ) \left (x^{2}+1\right )} \end {align*}
Hence the ode is \begin {align*} y^{\prime } = \frac {x \left (2 x +1\right )}{\left (1+x \right ) \left (x^{2}+1\right )} \end {align*}
The domain of \(p(x)=0\) is \[
\{-\infty
Integrating both sides gives \begin {align*} y &= \int { \frac {x \left (2 x +1\right )}{\left (1+x \right ) \left (x^{2}+1\right )}\,\mathop {\mathrm {d}x}}\\ &= \frac {\ln \left (1+x \right )}{2}+\frac {3 \ln \left (x^{2}+1\right )}{4}-\frac {\arctan \left (x \right )}{2}+c_{1} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(y=1\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} 1 = c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = 1 \end {align*}
Trying the constant \begin {align*} c_{1} = 1 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=\frac {\ln \left (1+x \right )}{2}+\frac {3 \ln \left (x^{2}+1\right )}{4}-\frac {\arctan \left (x \right )}{2}+1 \end {align*}
The constant \(c_{1} = 1\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {\ln \left (1+x \right )}{2}+\frac {3 \ln \left (x^{2}+1\right )}{4}-\frac {\arctan \left (x \right )}{2}+1 \\
\end{align*} Verification of solutions
\[
y = \frac {\ln \left (1+x \right )}{2}+\frac {3 \ln \left (x^{2}+1\right )}{4}-\frac {\arctan \left (x \right )}{2}+1
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\left (1+x \right ) \left (x^{2}+1\right ) y^{\prime }=2 x^{2}+x , y \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {2 x^{2}+x}{\left (1+x \right ) \left (x^{2}+1\right )} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int \frac {2 x^{2}+x}{\left (1+x \right ) \left (x^{2}+1\right )}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=\frac {\ln \left (1+x \right )}{2}+\frac {3 \ln \left (x^{2}+1\right )}{4}-\frac {\arctan \left (x \right )}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\ln \left (1+x \right )}{2}+\frac {3 \ln \left (x^{2}+1\right )}{4}-\frac {\arctan \left (x \right )}{2}+c_{1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =1 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =1\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\ln \left (1+x \right )}{2}+\frac {3 \ln \left (x^{2}+1\right )}{4}-\frac {\arctan \left (x \right )}{2}+1 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\ln \left (1+x \right )}{2}+\frac {3 \ln \left (x^{2}+1\right )}{4}-\frac {\arctan \left (x \right )}{2}+1 \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 24
\[
y \left (x \right ) = \frac {3 \ln \left (x^{2}+1\right )}{4}-\frac {\arctan \left (x \right )}{2}+\frac {\ln \left (x +1\right )}{2}+1
\]
✓ Solution by Mathematica
Time used: 0.01 (sec). Leaf size: 29
\[
y(x)\to \frac {1}{4} \left (-2 \arctan (x)+3 \log \left (x^2+1\right )+2 \log (x+1)+4\right )
\]
1.30.2 Solving as quadrature ode
1.30.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
<- quadrature successful`
dsolve([(x+1)*(x^2+1)*diff(y(x),x)=2*x^2+x,y(0) = 1],y(x), singsol=all)
DSolve[{(x+1)*(x^2+1)*y'[x]==2*x^2+x,{y[0]==1}},y[x],x,IncludeSingularSolutions -> True]