problem 3

Internal problem ID [6155]
Internal ﬁle name [OUTPUT/5403_Sunday_June_05_2022_03_36_23_PM_21409840/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 1.3 SEPARABLE EQUATIONS. Page 12
Problem number: 3.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_integrable_as_is", "second_order_ode_missing_y", "exact nonlinear second order ode"

2.17.1 Solving as second order integrable as is ode

Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int -\frac {y^{\prime \prime }}{y^{\prime }}d x &= \int -x^{2}d x\\ -\ln \left (y^{\prime }\right ) = -\frac {x^{3}}{3} + c_{1} \end {align*}

Which is now solved for \(y\). Integrating both sides gives \begin {align*} y &= \int { {\mathrm e}^{\frac {x^{3}}{3}-c_{1}}\,\mathop {\mathrm {d}x}}\\ &= -\frac {{\mathrm e}^{-c_{1}} 3^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} \left (\frac {2 x \sqrt {3}\, \left (-1\right )^{\frac {1}{3}} \pi }{3 \Gamma \left (\frac {2}{3}\right ) \left (-x^{3}\right )^{\frac {1}{3}}}-\frac {x \left (-1\right )^{\frac {1}{3}} \Gamma \left (\frac {1}{3}, -\frac {x^{3}}{3}\right )}{\left (-x^{3}\right )^{\frac {1}{3}}}\right )}{3}+c_{2} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {{\mathrm e}^{-c_{1}} 3^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} \left (\frac {2 x \sqrt {3}\, \left (-1\right )^{\frac {1}{3}} \pi }{3 \Gamma \left (\frac {2}{3}\right ) \left (-x^{3}\right )^{\frac {1}{3}}}-\frac {x \left (-1\right )^{\frac {1}{3}} \Gamma \left (\frac {1}{3}, -\frac {x^{3}}{3}\right )}{\left (-x^{3}\right )^{\frac {1}{3}}}\right )}{3}+c_{2} \\ \end{align*}

Veriﬁcation of solutions

\[ y = -\frac {{\mathrm e}^{-c_{1}} 3^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} \left (\frac {2 x \sqrt {3}\, \left (-1\right )^{\frac {1}{3}} \pi }{3 \Gamma \left (\frac {2}{3}\right ) \left (-x^{3}\right )^{\frac {1}{3}}}-\frac {x \left (-1\right )^{\frac {1}{3}} \Gamma \left (\frac {1}{3}, -\frac {x^{3}}{3}\right )}{\left (-x^{3}\right )^{\frac {1}{3}}}\right )}{3}+c_{2} \] Veriﬁed OK.

2.17.2 Solving as second order ode missing y ode

This is second order ode with missing dependent variable \(y\). Let \begin {align*} p(x) &= y^{\prime } \end {align*}

Hence the ode becomes \begin {align*} -\frac {p^{\prime }\left (x \right )}{p \left (x \right )}+x^{2} = 0 \end {align*}

Which is now solve for \(p(x)\) as ﬁrst order ode. In canonical form the ODE is \begin {align*} p' &= F(x,p)\\ &= f( x) g(p)\\ &= x^{2} p \end {align*}

Where \(f(x)=x^{2}\) and \(g(p)=p\). Integrating both sides gives \begin {align*} \frac {1}{p} \,dp &= x^{2} \,d x\\ \int { \frac {1}{p} \,dp} &= \int {x^{2} \,d x}\\ \ln \left (p \right )&=\frac {x^{3}}{3}+c_{1}\\ p&={\mathrm e}^{\frac {x^{3}}{3}+c_{1}}\\ &=c_{1} {\mathrm e}^{\frac {x^{3}}{3}} \end {align*}

Since \(p=y^{\prime }\) then the new ﬁrst order ode to solve is \begin {align*} y^{\prime } = c_{1} {\mathrm e}^{\frac {x^{3}}{3}} \end {align*}

Integrating both sides gives \begin {align*} y &= \int { c_{1} {\mathrm e}^{\frac {x^{3}}{3}}\,\mathop {\mathrm {d}x}}\\ &= -\frac {c_{1} 3^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} \left (\frac {2 x \sqrt {3}\, \left (-1\right )^{\frac {1}{3}} \pi }{3 \Gamma \left (\frac {2}{3}\right ) \left (-x^{3}\right )^{\frac {1}{3}}}-\frac {x \left (-1\right )^{\frac {1}{3}} \Gamma \left (\frac {1}{3}, -\frac {x^{3}}{3}\right )}{\left (-x^{3}\right )^{\frac {1}{3}}}\right )}{3}+c_{2} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {c_{1} 3^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} \left (\frac {2 x \sqrt {3}\, \left (-1\right )^{\frac {1}{3}} \pi }{3 \Gamma \left (\frac {2}{3}\right ) \left (-x^{3}\right )^{\frac {1}{3}}}-\frac {x \left (-1\right )^{\frac {1}{3}} \Gamma \left (\frac {1}{3}, -\frac {x^{3}}{3}\right )}{\left (-x^{3}\right )^{\frac {1}{3}}}\right )}{3}+c_{2} \\ \end{align*}

Veriﬁcation of solutions

\[ y = -\frac {c_{1} 3^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} \left (\frac {2 x \sqrt {3}\, \left (-1\right )^{\frac {1}{3}} \pi }{3 \Gamma \left (\frac {2}{3}\right ) \left (-x^{3}\right )^{\frac {1}{3}}}-\frac {x \left (-1\right )^{\frac {1}{3}} \Gamma \left (\frac {1}{3}, -\frac {x^{3}}{3}\right )}{\left (-x^{3}\right )^{\frac {1}{3}}}\right )}{3}+c_{2} \] Veriﬁed OK.

2.17.3 Solving as exact nonlinear second order ode ode

An exact non-linear second order ode has the form \begin {align*} a_{2} \left (x , y, y^{\prime }\right ) y^{\prime \prime }+a_{1} \left (x , y, y^{\prime }\right ) y^{\prime }+a_{0} \left (x , y, y^{\prime }\right )&=0 \end {align*}

Where the following conditions are satisﬁed \begin {align*} \frac {\partial a_2}{\partial y} &= \frac {\partial a_1}{\partial y'}\\ \frac {\partial a_2}{\partial x} &= \frac {\partial a_0}{\partial y'}\\ \frac {\partial a_1}{\partial x} &= \frac {\partial a_0}{\partial y} \end {align*}

Looking at the the ode given we see that \begin {align*} a_2 &= -\frac {1}{y^{\prime }}\\ a_1 &= 0\\ a_0 &= x^{2} \end {align*}

Applying the conditions to the above shows this is a nonlinear exact second order ode. Therefore it can be reduced to ﬁrst order ode given by \begin {align*} \int {a_2\,d y'} + \int {a_1\,d y} + \int {a_0\,d x} &= c_{1}\\ \int {-\frac {1}{y^{\prime }}\,d y'} + \int {0\,d y} + \int {x^{2}\,d x} &= c_{1} \end {align*}

Which results in \begin {align*} \frac {x^{3}}{3}-\ln \left (y^{\prime }\right ) = c_{1} \end {align*}

Which is now solved Integrating both sides gives \begin {align*} y &= \int { {\mathrm e}^{\frac {x^{3}}{3}-c_{1}}\,\mathop {\mathrm {d}x}}\\ &= -\frac {{\mathrm e}^{-c_{1}} 3^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} \left (\frac {2 x \sqrt {3}\, \left (-1\right )^{\frac {1}{3}} \pi }{3 \Gamma \left (\frac {2}{3}\right ) \left (-x^{3}\right )^{\frac {1}{3}}}-\frac {x \left (-1\right )^{\frac {1}{3}} \Gamma \left (\frac {1}{3}, -\frac {x^{3}}{3}\right )}{\left (-x^{3}\right )^{\frac {1}{3}}}\right )}{3}+c_{2} \end {align*}

Summary

Veriﬁcation of solutions

2.17.4 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -\frac {\frac {d}{d x}y^{\prime }}{y^{\prime }}=-x^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=y^{\prime } x^{2} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & -y^{\prime } x^{2}+\frac {d}{d x}y^{\prime }=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} k \,x^{k +1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -1 \\ {} & {} & x^{2}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} \left (k -1\right ) x^{k} \\ {} & \circ & \textrm {Convert}\hspace {3pt} \frac {d}{d x}y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\moverset {\infty }{\munderset {k =2}{\sum }}a_{k} k \left (k -1\right ) x^{k -2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \frac {d}{d x}y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k +2} \left (k +2\right ) \left (k +1\right ) x^{k} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 2 a_{2}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +2} \left (k +2\right ) \left (k +1\right )-a_{k -1} \left (k -1\right )\right ) x^{k}\right )=0 \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 2 a_{2}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k^{2}+3 k +2\right ) a_{k +2}-a_{k -1} \left (k -1\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (\left (k +1\right )^{2}+3 k +5\right ) a_{k +3}-a_{k} k =0 \\ \bullet & {} & \textrm {Recursion relation that defines the series solution to the ODE}\hspace {3pt} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +3}=\frac {a_{k} k}{k^{2}+5 k +6}, 2 a_{2}=0\right ] \end {array} \]

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
checking if the LODE is of Euler type 
trying a symmetry of the form [xi=0, eta=F(x)] 
checking if the LODE is missing y 
<- LODE missing y successful`

\[ y \left (x \right ) = -3 \Gamma \left (\frac {1}{3}, -\frac {x^{3}}{3}\right ) c_{2} \Gamma \left (\frac {2}{3}\right )+2 \sqrt {3}\, \pi c_{2} +c_{1} \]

\[ y(x)\to \frac {c_1 \left (-x^3\right )^{2/3} \Gamma \left (\frac {1}{3},-\frac {x^3}{3}\right )}{3^{2/3} x^2}+c_2 \]

2.17 problem 3

2.17.1 Solving as second order integrable as is ode

2.17.2 Solving as second order ode missing y ode

2.17.3 Solving as exact nonlinear second order ode ode

2.17.4 Maple step by step solution