Internal problem ID [2684]
Internal file name [OUTPUT/2176_Sunday_June_05_2022_02_52_06_AM_56666401/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 29(a).
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 A`]]
\[ \boxed {y^{\prime }-\frac {x +y a}{a x -y}=0} \]
In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {y a +x}{-a 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 \(M=y a +x\) and \(N=a x -y\) 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 {a u +1}{a -u}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {\frac {a u \left (x \right )+1}{a -u \left (x \right )}-u \left (x \right )}{x} \end {align*}
Or \[ u^{\prime }\left (x \right )-\frac {\frac {a u \left (x \right )+1}{a -u \left (x \right )}-u \left (x \right )}{x} = 0 \] Or \[ u^{\prime }\left (x \right ) x u \left (x \right )-u^{\prime }\left (x \right ) x a +u \left (x \right )^{2}+1 = 0 \] Or \[ -x \left (a -u \left (x \right )\right ) u^{\prime }\left (x \right )+u \left (x \right )^{2}+1 = 0 \] 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 {-u^{2}-1}{x \left (a -u \right )} \end {align*}
Where \(f(x)=-\frac {1}{x}\) and \(g(u)=\frac {-u^{2}-1}{a -u}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {-u^{2}-1}{a -u}} \,du &= -\frac {1}{x} \,d x \\ \int { \frac {1}{\frac {-u^{2}-1}{a -u}} \,du} &= \int {-\frac {1}{x} \,d x} \\ \frac {\ln \left (u^{2}+1\right )}{2}-a \arctan \left (u \right )&=-\ln \left (x \right )+c_{2} \\ \end{align*} The solution is \[ \frac {\ln \left (u \left (x \right )^{2}+1\right )}{2}-a \arctan \left (u \left (x \right )\right )+\ln \left (x \right )-c_{2} = 0 \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ \frac {\ln \left (\frac {y^{2}}{x^{2}}+1\right )}{2}-a \arctan \left (\frac {y}{x}\right )+\ln \left (x \right )-c_{2} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} \frac {\ln \left (\frac {y^{2}}{x^{2}}+1\right )}{2}-a \arctan \left (\frac {y}{x}\right )+\ln \left (x \right )-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {\ln \left (\frac {y^{2}}{x^{2}}+1\right )}{2}-a \arctan \left (\frac {y}{x}\right )+\ln \left (x \right )-c_{2} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {x +y a}{a x -y}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x +y a}{a x -y} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying homogeneous D <- homogeneous successful`
✓ Solution by Maple
Time used: 0.359 (sec). Leaf size: 25
dsolve(diff(y(x),x)=(x+a*y(x))/(a*x-y(x)),y(x), singsol=all)
\[ y \left (x \right ) = \tan \left (\operatorname {RootOf}\left (-2 a \textit {\_Z} +\ln \left (\sec \left (\textit {\_Z} \right )^{2} x^{2}\right )+2 c_{1} \right )\right ) x \]
✓ Solution by Mathematica
Time used: 0.039 (sec). Leaf size: 34
DSolve[y'[x]==(x+a*y[x])/(a*x-y[x]),y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [a \arctan \left (\frac {y(x)}{x}\right )-\frac {1}{2} \log \left (\frac {y(x)^2}{x^2}+1\right )=\log (x)+c_1,y(x)\right ] \]