Internal problem ID [2677]
Internal file name [OUTPUT/2169_Sunday_June_05_2022_02_51_33_AM_71672999/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 21.
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`], _rational, [_Abel, `2nd type`, `class B`]]
\[ \boxed {2 x \left (y+2 x \right ) y^{\prime }-y \left (4 x -y\right )=0} \]
In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {y \left (-4 x +y \right )}{2 x \left (y +2 x \right )}\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 \(M=y \left (4 x -y \right )\) and \(N=2 x \left (y +2 x \right )\) are both homogeneous and of the same order \(n=2\). 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 {u \left (-4+u \right )}{2 u +4}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {-\frac {u \left (x \right ) \left (-4+u \left (x \right )\right )}{2 u \left (x \right )+4}-u \left (x \right )}{x} \end {align*}
Or \[ u^{\prime }\left (x \right )-\frac {-\frac {u \left (x \right ) \left (-4+u \left (x \right )\right )}{2 u \left (x \right )+4}-u \left (x \right )}{x} = 0 \] Or \[ 2 u^{\prime }\left (x \right ) x u \left (x \right )+4 u^{\prime }\left (x \right ) x +3 u \left (x \right )^{2} = 0 \] Or \[ 2 x \left (u \left (x \right )+2\right ) u^{\prime }\left (x \right )+3 u \left (x \right )^{2} = 0 \] 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 {3 u^{2}}{2 x \left (u +2\right )} \end {align*}
Where \(f(x)=-\frac {3}{2 x}\) and \(g(u)=\frac {u^{2}}{u +2}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{2}}{u +2}} \,du &= -\frac {3}{2 x} \,d x \\ \int { \frac {1}{\frac {u^{2}}{u +2}} \,du} &= \int {-\frac {3}{2 x} \,d x} \\ -\frac {2}{u}+\ln \left (u \right )&=-\frac {3 \ln \left (x \right )}{2}+c_{2} \\ \end{align*} The solution is \[ -\frac {2}{u \left (x \right )}+\ln \left (u \left (x \right )\right )+\frac {3 \ln \left (x \right )}{2}-c_{2} = 0 \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ \ln \left (\frac {y}{x}\right )-\frac {2 x}{y}+\frac {3 \ln \left (x \right )}{2}-c_{2} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} \ln \left (\frac {y}{x}\right )-\frac {2 x}{y}+\frac {3 \ln \left (x \right )}{2}-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \ln \left (\frac {y}{x}\right )-\frac {2 x}{y}+\frac {3 \ln \left (x \right )}{2}-c_{2} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x \left (y+2 x \right ) y^{\prime }-y \left (4 x -y\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y \left (4 x -y\right )}{2 x \left (y+2 x \right )} \end {array} \]
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: 0.0 (sec). Leaf size: 19
dsolve(2*x*(y(x)+2*x)*diff(y(x),x)=y(x)*(4*x-y(x)),y(x), singsol=all)
\[ y \left (x \right ) = \frac {2 x}{\operatorname {LambertW}\left (2 \,{\mathrm e}^{\frac {3 c_{1}}{2}} x^{\frac {3}{2}}\right )} \]
✓ Solution by Mathematica
Time used: 5.346 (sec). Leaf size: 29
DSolve[2*x*(y[x]+2*x)*y'[x]==y[x]*(4*x-y[x]),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {2 x}{W\left (2 e^{-c_1} x^{3/2}\right )} \\ y(x)\to 0 \\ \end{align*}