Internal problem ID [6207]
Internal file name [OUTPUT/5455_Sunday_June_05_2022_03_39_07_PM_87586332/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 1.7. Homogeneous Equations. Page
28
Problem number: 1(e).
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program :
Maple gives the following as the ode type
[[_homogeneous, `class A`], _dAlembert]
\[ \boxed {y^{\prime } x -y-2 x \,{\mathrm e}^{-\frac {y}{x}}=0} \]
In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {y +2 x \,{\mathrm e}^{-\frac {y}{x}}}{x}\tag {1} \end {align*}
An ode of the form \(y' = \frac {M(x,y)}{N(x,y)}\) is called homogeneous if the functions \(M(x,y)\) and \(N(x,y)\) are both homogeneous functions and of the same order. Recall that a function \(f(x,y)\) is homogeneous of order \(n\) if \[ f(t^n x, t^n y)= t^n f(x,y) \] In this case, it can be seen that both \(M=y +2 x \,{\mathrm e}^{-\frac {y}{x}}\) and \(N=x\) are both homogeneous and of the same order \(n=1\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {y}{x}\), or \(y=ux\). Hence \[ \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u \] Applying the transformation \(y=ux\) to the above ODE in (1) gives \begin {align*} \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u &= u +2 \,{\mathrm e}^{-u}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {2 \,{\mathrm e}^{-u \left (x \right )}}{x} \end {align*}
Or \[ u^{\prime }\left (x \right )-\frac {2 \,{\mathrm e}^{-u \left (x \right )}}{x} = 0 \] Or \[ u^{\prime }\left (x \right ) {\mathrm e}^{u \left (x \right )} x -2 = 0 \] Which is now solved as separable in \(u \left (x \right )\). Which is now solved in \(u \left (x \right )\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {2 \,{\mathrm e}^{-u}}{x} \end {align*}
Where \(f(x)=\frac {2}{x}\) and \(g(u)={\mathrm e}^{-u}\). Integrating both sides gives \begin{align*} \frac {1}{{\mathrm e}^{-u}} \,du &= \frac {2}{x} \,d x \\ \int { \frac {1}{{\mathrm e}^{-u}} \,du} &= \int {\frac {2}{x} \,d x} \\ {\mathrm e}^{u}&=2 \ln \left (x \right )+c_{2} \\ \end{align*} The solution is \[ {\mathrm e}^{u \left (x \right )}-2 \ln \left (x \right )-c_{2} = 0 \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ {\mathrm e}^{\frac {y}{x}}-2 \ln \left (x \right )-c_{2} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} {\mathrm e}^{\frac {y}{x}}-2 \ln \left (x \right )-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ {\mathrm e}^{\frac {y}{x}}-2 \ln \left (x \right )-c_{2} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x -y-2 x \,{\mathrm e}^{-\frac {y}{x}}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y+2 x \,{\mathrm e}^{-\frac {y}{x}}}{x} \end {array} \]
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 homogeneous D <- homogeneous successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 14
dsolve(x*diff(y(x),x)=y(x)+2*x*exp(-y(x)/x),y(x), singsol=all)
\[ y \left (x \right ) = \left (\ln \left (2\right )+\ln \left (\ln \left (x \right )+c_{1} \right )\right ) x \]
✓ Solution by Mathematica
Time used: 0.42 (sec). Leaf size: 15
DSolve[x*y'[x]==y[x]+2*x*Exp[-y[x]/x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to x \log (2 \log (x)+c_1) \]