problem Problem 10

Internal problem ID [2666]
Internal ﬁle name [OUTPUT/2158_Sunday_June_05_2022_02_50_52_AM_39559977/index.tex]

Book: Diﬀerential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section: Chapter 1, First-Order Diﬀerential Equations. Section 1.8, Change of Variables. page 79
Problem number: Problem 10.
ODE order: 1.
ODE degree: 1.

4.2.1 Solving as homogeneous ode

In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {3 y}{-3 x +y}\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 \(M=3 y\) and \(N=3 x -y\) 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 &= -\frac {3 u}{u -3}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {-\frac {3 u \left (x \right )}{u \left (x \right )-3}-u \left (x \right )}{x} \end {align*}

Or \[ u^{\prime }\left (x \right )-\frac {-\frac {3 u \left (x \right )}{u \left (x \right )-3}-u \left (x \right )}{x} = 0 \] Or \[ u^{\prime }\left (x \right ) x u \left (x \right )-3 u^{\prime }\left (x \right ) x +u \left (x \right )^{2} = 0 \] Or \[ x \left (u \left (x \right )-3\right ) u^{\prime }\left (x \right )+u \left (x \right )^{2} = 0 \] 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 {u^{2}}{x \left (u -3\right )} \end {align*}

Where \(f(x)=-\frac {1}{x}\) and \(g(u)=\frac {u^{2}}{u -3}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{2}}{u -3}} \,du &= -\frac {1}{x} \,d x \\ \int { \frac {1}{\frac {u^{2}}{u -3}} \,du} &= \int {-\frac {1}{x} \,d x} \\ \frac {3}{u}+\ln \left (u \right )&=-\ln \left (x \right )+c_{2} \\ \end{align*} The solution is \[ \frac {3}{u \left (x \right )}+\ln \left (u \left (x \right )\right )+\ln \left (x \right )-c_{2} = 0 \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ \ln \left (\frac {y}{x}\right )+\frac {3 x}{y}+\ln \left (x \right )-c_{2} = 0 \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} \ln \left (\frac {y}{x}\right )+\frac {3 x}{y}+\ln \left (x \right )-c_{2} &= 0 \\ \end{align*}

Veriﬁcation of solutions

\[ \ln \left (\frac {y}{x}\right )+\frac {3 x}{y}+\ln \left (x \right )-c_{2} = 0 \] Veriﬁed OK.

4.2.2 Maple step by step solution

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

Maple trace

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

\[ y \left (x \right ) = -\frac {3 x}{\operatorname {LambertW}\left (-3 x \,{\mathrm e}^{-3 c_{1}}\right )} \]

\begin{align*} y(x)\to -\frac {3 x}{W\left (-3 e^{-c_1} x\right )} \\ y(x)\to 0 \\ \end{align*}

4.2 problem Problem 10

4.2.1 Solving as homogeneous ode

4.2.2 Maple step by step solution