1.432 problem 434

1.432.1 Solving as quadrature ode
1.432.2 Maple step by step solution

Internal problem ID [8769]
Internal file name [OUTPUT/7703_Sunday_June_05_2022_11_44_06_PM_22127683/index.tex]

Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 434.
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]

\[ \boxed {y^{\prime }=1} \]

1.432.1 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 452: Slope field plot

Verification of solutions

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

1.432.2 Maple step by step solution

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

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
<- 1st order linear successful`
 

Solution by Maple

Time used: 0.0 (sec). Leaf size: 7

dsolve(diff(y(x),x)-1 = 0,y(x), singsol=all)
 

\[ y \left (x \right ) = x +c_{1} \]

Solution by Mathematica

Time used: 0.116 (sec). Leaf size: 71

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

\begin{align*} y(x)\to \frac {1}{2} e^{-c_1} \left (-x^2+e^{2 c_1}\right ) \\ y(x)\to \frac {1}{2} e^{-c_1} \left (-1+e^{2 c_1} x^2\right ) \\ y(x)\to -i x \\ y(x)\to i x \\ \end{align*}