problem 39 part(a)

Internal problem ID [1149]
Internal ﬁle name [OUTPUT/1150_Sunday_June_05_2022_02_03_35_AM_37748633/index.tex]

Book: Elementary diﬀerential equations with boundary value problems. William F. Trench. Brooks/Cole 2001
Section: Chapter 5 linear second order equations. Section 5.6 Reduction or order. Page 253
Problem number: 39 part(a).
ODE order: 1.
ODE degree: 1.

\[ \boxed {x^{2} \left (y^{\prime }+y^{2}\right )-x \left (2+x \right ) y=-x -2} \]

9.43.1 Solving as riccati ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {x^{2} y^{2}-x^{2} y -2 y x +x +2}{x^{2}} \end {align*}

This is a Riccati ODE. Comparing the ODE to solve \[ y' = -y^{2}+y +\frac {2 y}{x}-\frac {1}{x}-\frac {2}{x^{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-\frac {2+x}{x^{2}}\), \(f_1(x)=-\frac {-x^{2}-2 x}{x^{2}}\) 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 &=\frac {-x^{2}-2 x}{x^{2}}\\ f_2^2 f_0 &=-\frac {2+x}{x^{2}} \end {align*}

Substituting the above terms back in equation (2) gives \begin {align*} -u^{\prime \prime }\left (x \right )-\frac {\left (-x^{2}-2 x \right ) u^{\prime }\left (x \right )}{x^{2}}-\frac {\left (2+x \right ) u \left (x \right )}{x^{2}} &=0 \end {align*}

\[ u \left (x \right ) = x \left (c_{2} {\mathrm e}^{x}+c_{1} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \left (x +1\right ) c_{2} {\mathrm e}^{x}+c_{1} \] Using the above in (1) gives the solution \[ y = \frac {\left (x +1\right ) c_{2} {\mathrm e}^{x}+c_{1}}{x \left (c_{2} {\mathrm e}^{x}+c_{1} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution

\[ y = \frac {x \,{\mathrm e}^{x}+{\mathrm e}^{x}+c_{3}}{x \left ({\mathrm e}^{x}+c_{3} \right )} \]

Summary

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

Veriﬁcation of solutions

\[ y = \frac {x \,{\mathrm e}^{x}+{\mathrm e}^{x}+c_{3}}{x \left ({\mathrm e}^{x}+c_{3} \right )} \] Veriﬁed OK.

9.43.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (y^{\prime }+y^{2}\right )-x \left (2+x \right ) y=-x -2 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {x^{2} y^{2}-x^{2} y-2 y x +x +2}{x^{2}} \end {array} \]

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 
found: 2 potential symmetries. Proceeding with integration step`

\[ y \left (x \right ) = \frac {x \,{\mathrm e}^{x}+{\mathrm e}^{x}-c_{1}}{\left (-c_{1} +{\mathrm e}^{x}\right ) x} \]

\[ y(x)\to e^{\frac {1}{x}-x} \left (\int _1^x\frac {e^{K[1]-\frac {1}{K[1]}} \left (K[1]^2+K[1]-2\right )}{K[1]^2}dK[1]+c_1\right ) \]

9.43 problem 39 part(a)

9.43.1 Solving as riccati ode

9.43.2 Maple step by step solution