Internal problem ID [11970]
Internal file name [OUTPUT/10623_Saturday_September_02_2023_02_48_36_PM_76338301/index.tex]
Book: AN INTRODUCTION TO ORDINARY DIFFERENTIAL EQUATIONS by JAMES C.
ROBINSON. Cambridge University Press 2004
Section: Chapter 5, Trivial differential equations. Exercises page 33
Problem number: 5.1 (iv).
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]
\[ \boxed {z^{\prime }=x \,{\mathrm e}^{-2 x}} \]
Integrating both sides gives \begin {align*} z &= \int { x \,{\mathrm e}^{-2 x}\,\mathop {\mathrm {d}x}}\\ &= \left (-\frac {x}{2}-\frac {1}{4}\right ) {\mathrm e}^{-2 x}+c_{1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} z &= \left (-\frac {x}{2}-\frac {1}{4}\right ) {\mathrm e}^{-2 x}+c_{1} \\ \end{align*}
Verification of solutions
\[ z = \left (-\frac {x}{2}-\frac {1}{4}\right ) {\mathrm e}^{-2 x}+c_{1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & z^{\prime }=x \,{\mathrm e}^{-2 x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & z^{\prime } \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int z^{\prime }d x =\int x \,{\mathrm e}^{-2 x}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & z=-\frac {\left (2 x +1\right ) {\mathrm e}^{-2 x}}{4}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} z \\ {} & {} & z=-\frac {x \,{\mathrm e}^{-2 x}}{2}-\frac {{\mathrm e}^{-2 x}}{4}+c_{1} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature <- quadrature successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 17
dsolve(diff(z(x),x)=x*exp(-2*x),z(x), singsol=all)
\[ z \left (x \right ) = \frac {\left (-2 x -1\right ) {\mathrm e}^{-2 x}}{4}+c_{1} \]
✓ Solution by Mathematica
Time used: 0.009 (sec). Leaf size: 22
DSolve[z'[x]==x*Exp[-2*x],z[x],x,IncludeSingularSolutions -> True]
\[ z(x)\to -\frac {1}{4} e^{-2 x} (2 x+1)+c_1 \]