Internal problem ID [2676]
Internal file name [OUTPUT/2168_Sunday_June_05_2022_02_51_28_AM_49626620/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 20.
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, _dAlembert]
\[ \boxed {y y^{\prime }-\sqrt {x^{2}+y^{2}}=-x} \]
In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {\sqrt {x^{2}+y^{2}}-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 \(M=\sqrt {x^{2}+y^{2}}-x\) and \(N=y\) 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 {\sqrt {u^{2}+1}-1}{u}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {\frac {\sqrt {u \left (x \right )^{2}+1}-1}{u \left (x \right )}-u \left (x \right )}{x} \end {align*}
Or \[ u^{\prime }\left (x \right )-\frac {\frac {\sqrt {u \left (x \right )^{2}+1}-1}{u \left (x \right )}-u \left (x \right )}{x} = 0 \] Or \[ u^{\prime }\left (x \right ) u \left (x \right ) x +u \left (x \right )^{2}-\sqrt {u \left (x \right )^{2}+1}+1 = 0 \] 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}-\sqrt {u^{2}+1}+1}{u x} \end {align*}
Where \(f(x)=-\frac {1}{x}\) and \(g(u)=\frac {u^{2}-\sqrt {u^{2}+1}+1}{u}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{2}-\sqrt {u^{2}+1}+1}{u}} \,du &= -\frac {1}{x} \,d x \\ \int { \frac {1}{\frac {u^{2}-\sqrt {u^{2}+1}+1}{u}} \,du} &= \int {-\frac {1}{x} \,d x} \\ \ln \left (\sqrt {u^{2}+1}-1\right )&=-\ln \left (x \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {u^{2}+1}-1 &= {\mathrm e}^{-\ln \left (x \right )+c_{2}} \end {align*}
Which simplifies to \begin {align*} \sqrt {u^{2}+1}-1 &= \frac {c_{3}}{x} \end {align*}
Which simplifies to \[ \sqrt {u \left (x \right )^{2}+1}-1 = \frac {c_{3} {\mathrm e}^{c_{2}}}{x} \] The solution is \[ \sqrt {u \left (x \right )^{2}+1}-1 = \frac {c_{3} {\mathrm e}^{c_{2}}}{x} \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ \sqrt {\frac {y^{2}}{x^{2}}+1}-1 = \frac {c_{3} {\mathrm e}^{c_{2}}}{x} \]
The solution(s) found are the following \begin{align*} \tag{1} \sqrt {\frac {x^{2}+y^{2}}{x^{2}}}-1 &= \frac {c_{3} {\mathrm e}^{c_{2}}}{x} \\ \end{align*}
Verification of solutions
\[
\sqrt {\frac {x^{2}+y^{2}}{x^{2}}}-1 = \frac {c_{3} {\mathrm e}^{c_{2}}}{x}
\] Verified OK. {0 < x}
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y y^{\prime }-\sqrt {x^{2}+y^{2}}=-x \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\sqrt {x^{2}+y^{2}}-x}{y} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying homogeneous types: trying homogeneous C 1st order, trying the canonical coordinates of the invariance group <- 1st order, canonical coordinates successful <- homogeneous successful`
✓ Solution by Maple
Time used: 0.062 (sec). Leaf size: 27
dsolve(y(x)*diff(y(x),x)=sqrt(x^2+y(x)^2)-x,y(x), singsol=all)
\[ \frac {-c_{1} y \left (x \right )^{2}+\sqrt {x^{2}+y \left (x \right )^{2}}+x}{y \left (x \right )^{2}} = 0 \]
✓ Solution by Mathematica
Time used: 0.409 (sec). Leaf size: 57
DSolve[y[x]*y'[x]==Sqrt[x^2+y[x]^2]-x,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -e^{\frac {c_1}{2}} \sqrt {2 x+e^{c_1}} \\ y(x)\to e^{\frac {c_1}{2}} \sqrt {2 x+e^{c_1}} \\ y(x)\to 0 \\ \end{align*}