problem 4(d)

Internal problem ID [6074]
Internal ﬁle name [OUTPUT/5322_Sunday_June_05_2022_03_34_09_PM_77661497/index.tex]

Book: An introduction to Ordinary Diﬀerential Equations. Earl A. Coddington. Dover. NY 1961
Section: Chapter 5. Existence and uniqueness of solutions to ﬁrst order equations. Page 190
Problem number: 4(d).
ODE order: 1.
ODE degree: 1.

21.12.1 Solving as homogeneous ode

In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {y +x \,{\mathrm e}^{-\frac {2 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 \(M=y +x \,{\mathrm e}^{-\frac {2 y}{x}}\) 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 +{\mathrm e}^{-2 u}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {{\mathrm e}^{-2 u \left (x \right )}}{x} \end {align*}

Or \[ u^{\prime }\left (x \right )-\frac {{\mathrm e}^{-2 u \left (x \right )}}{x} = 0 \] Or \[ u^{\prime }\left (x \right ) {\mathrm e}^{2 u \left (x \right )} x -1 = 0 \] Or \[ u^{\prime }\left (x \right ) {\mathrm e}^{2 u \left (x \right )} x -1 = 0 \] 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 {{\mathrm e}^{-2 u}}{x} \end {align*}

Where \(f(x)=\frac {1}{x}\) and \(g(u)={\mathrm e}^{-2 u}\). Integrating both sides gives \begin{align*} \frac {1}{{\mathrm e}^{-2 u}} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{{\mathrm e}^{-2 u}} \,du} &= \int {\frac {1}{x} \,d x} \\ \frac {{\mathrm e}^{2 u}}{2}&=\ln \left (x \right )+c_{2} \\ \end{align*} The solution is \[ \frac {{\mathrm e}^{2 u \left (x \right )}}{2}-\ln \left (x \right )-c_{2} = 0 \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ \frac {{\mathrm e}^{\frac {2 y}{x}}}{2}-\ln \left (x \right )-c_{2} = 0 \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {{\mathrm e}^{\frac {2 y}{x}}}{2}-\ln \left (x \right )-c_{2} &= 0 \\ \end{align*}

Veriﬁcation of solutions

\[ \frac {{\mathrm e}^{\frac {2 y}{x}}}{2}-\ln \left (x \right )-c_{2} = 0 \] Veriﬁed OK.

21.12.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {y+x \,{\mathrm e}^{-\frac {2 y}{x}}}{x}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y+x \,{\mathrm e}^{-\frac {2 y}{x}}}{x} \end {array} \]

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`

\[ y \left (x \right ) = \frac {\left (\ln \left (2\right )+\ln \left (\ln \left (x \right )+c_{1} \right )\right ) x}{2} \]

21.12 problem 4(d)

21.12.1 Solving as homogeneous ode

21.12.2 Maple step by step solution