6.5 problem 5

6.5.1 Solving as quadrature ode
6.5.2 Maple step by step solution

Internal problem ID [1034]
Internal file name [OUTPUT/1035_Sunday_June_05_2022_01_57_26_AM_68709905/index.tex]

Book: Elementary differential equations with boundary value problems. William F. Trench. Brooks/Cole 2001
Section: Chapter 2, First order equations. Exact equations. Section 2.5 Page 79
Problem number: 5.
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]

\[ \boxed {\left (x +y\right )^{2}+\left (x +y\right )^{2} y^{\prime }=0} \]

6.5.1 Solving as quadrature ode

Integrating both sides gives \begin {align*} y &= \int { -1\,\mathop {\mathrm {d}x}}\\ &= -x +c_{1} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -x +c_{1} \\ \end{align*}

Figure 251: Slope field plot

Verification of solutions

\[ y = -x +c_{1} \] Verified OK.

6.5.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x +y\right )^{2}+\left (x +y\right )^{2} y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & y^{\prime }=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int \left (-1\right )d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=-x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-x +c_{1} \end {array} \]

Maple trace

`Classification methods on request 
Methods to be used are: [exact] 
---------------------------- 
* Tackling ODE using method: exact 
--- Trying classification methods --- 
trying exact 
<- exact successful`
 

Solution by Maple

Time used: 0.0 (sec). Leaf size: 49

dsolve((x+y(x))^2+(x+y(x))^2*diff(y(x),x)=0,y(x), singsol=all)
 

\begin{align*} y \left (x \right ) &= -x \\ y \left (x \right ) &= c_{1} -x \\ y \left (x \right ) &= -\frac {c_{1}}{2}-\frac {i \sqrt {3}\, c_{1}}{2}-x \\ y \left (x \right ) &= -\frac {c_{1}}{2}+\frac {i \sqrt {3}\, c_{1}}{2}-x \\ \end{align*}

Solution by Mathematica

Time used: 0.002 (sec). Leaf size: 18

DSolve[(x+y[x])^2+(x+y[x])^2*y'[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to -x \\ y(x)\to -x+c_1 \\ \end{align*}