Internal problem ID [8462]
Internal file name [OUTPUT/7395_Sunday_June_05_2022_10_54_26_PM_65731826/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 126.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "exactWithIntegrationFactor", "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[[_homogeneous, `class G`]]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime } x -y f \left (y x \right )=0} \]
Writing the ode as \begin {align*} y^{\prime }&=\frac {y f \left (x y \right )}{x}\\ y^{\prime }&= \omega \left ( x,y\right ) \end {align*}
The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end {align*}
The type of this ode is not in the lookup table. To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives \begin{align*} \tag{1E} \xi &= x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x b_{2}+y b_{3}+b_{1} \\ \end{align*} Where the unknown coefficients are \[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \] Substituting equations (1E,2E) and \(\omega \) into (A) gives \begin{equation} \tag{5E} b_{2}+\frac {y f \left (x y \right ) \left (b_{3}-a_{2}\right )}{x}-\frac {y^{2} f \left (x y \right )^{2} a_{3}}{x^{2}}-\left (\frac {y^{2} D\left (f \right )\left (x y \right )}{x}-\frac {y f \left (x y \right )}{x^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {f \left (x y \right )}{x}+D\left (f \right )\left (x y \right ) y \right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation} Putting the above in normal form gives \[ -\frac {D\left (f \right )\left (x y \right ) x^{3} y b_{2}+D\left (f \right )\left (x y \right ) x^{2} y^{2} a_{2}+D\left (f \right )\left (x y \right ) x^{2} y^{2} b_{3}+D\left (f \right )\left (x y \right ) x \,y^{3} a_{3}+y^{2} f \left (x y \right )^{2} a_{3}+D\left (f \right )\left (x y \right ) x^{2} y b_{1}+D\left (f \right )\left (x y \right ) x \,y^{2} a_{1}+f \left (x y \right ) x^{2} b_{2}-f \left (x y \right ) y^{2} a_{3}+f \left (x y \right ) x b_{1}-f \left (x y \right ) y a_{1}-b_{2} x^{2}}{x^{2}} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} -D\left (f \right )\left (x y \right ) x^{3} y b_{2}-D\left (f \right )\left (x y \right ) x^{2} y^{2} a_{2}-D\left (f \right )\left (x y \right ) x^{2} y^{2} b_{3}-D\left (f \right )\left (x y \right ) x \,y^{3} a_{3}-y^{2} f \left (x y \right )^{2} a_{3}-D\left (f \right )\left (x y \right ) x^{2} y b_{1}-D\left (f \right )\left (x y \right ) x \,y^{2} a_{1}-f \left (x y \right ) x^{2} b_{2}+f \left (x y \right ) y^{2} a_{3}-f \left (x y \right ) x b_{1}+f \left (x y \right ) y a_{1}+b_{2} x^{2} = 0 \end{equation} Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \{x, y, f \left (x y \right ), D\left (f \right )\left (x y \right )\} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \{x = v_{1}, y = v_{2}, f \left (x y \right ) = v_{3}, D\left (f \right )\left (x y \right ) = v_{4}\} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} -v_{4} v_{1}^{2} v_{2}^{2} a_{2}-v_{4} v_{1} v_{2}^{3} a_{3}-v_{4} v_{1}^{3} v_{2} b_{2}-v_{4} v_{1}^{2} v_{2}^{2} b_{3}-v_{4} v_{1} v_{2}^{2} a_{1}-v_{2}^{2} v_{3}^{2} a_{3}-v_{4} v_{1}^{2} v_{2} b_{1}+v_{3} v_{2}^{2} a_{3}-v_{3} v_{1}^{2} b_{2}+v_{3} v_{2} a_{1}-v_{3} v_{1} b_{1}+b_{2} v_{1}^{2} = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}, v_{4}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} -v_{4} v_{1}^{3} v_{2} b_{2}+\left (-a_{2}-b_{3}\right ) v_{1}^{2} v_{2}^{2} v_{4}-v_{4} v_{1}^{2} v_{2} b_{1}-v_{3} v_{1}^{2} b_{2}+b_{2} v_{1}^{2}-v_{4} v_{1} v_{2}^{3} a_{3}-v_{4} v_{1} v_{2}^{2} a_{1}-v_{3} v_{1} b_{1}-v_{2}^{2} v_{3}^{2} a_{3}+v_{3} v_{2}^{2} a_{3}+v_{3} v_{2} a_{1} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} a_{1}&=0\\ a_{3}&=0\\ b_{2}&=0\\ -a_{1}&=0\\ -a_{3}&=0\\ -b_{1}&=0\\ -b_{2}&=0\\ -a_{2}-b_{3}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=-b_{3}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3} \end {align*}
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives \begin{align*} \xi &= -x \\ \eta &= y \\ \end{align*} The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is \begin {align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end {align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Unable to determine \(R\). Terminating
Entering Exact first order ODE solver. (Form one type)
To solve an ode of the form\begin {equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A} \end {equation} We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \] Hence\begin {equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B} \end {equation} Comparing (A,B) shows that\begin {align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end {align*}
But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\] If the above condition is satisfied, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is to write the ODE in standard form to check for exactness, which is \[ M(x,y) \mathop {\mathrm {d}x}+ N(x,y) \mathop {\mathrm {d}y}=0 \tag {1A} \] Therefore \begin {align*} \left (x\right )\mathop {\mathrm {d}y} &= \left (f \left (x y \right ) y\right )\mathop {\mathrm {d}x}\\ \left (-f \left (x y \right ) y\right )\mathop {\mathrm {d}x} + \left (x\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end {align*}
Comparing (1A) and (2A) shows that \begin {align*} M(x,y) &= -f \left (x y \right ) y\\ N(x,y) &= x \end {align*}
The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied \[ \frac {\partial M}{\partial y} = \frac {\partial N}{\partial x} \] Using result found above gives \begin {align*} \frac {\partial M}{\partial y} &= \frac {\partial }{\partial y} \left (-f \left (x y \right ) y\right )\\ &= -D\left (f \right )\left (x y \right ) x y -f \left (x y \right ) \end {align*}
And \begin {align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (x\right )\\ &= 1 \end {align*}
Since \(\frac {\partial M}{\partial y} \neq \frac {\partial N}{\partial x}\), then the ODE is not exact. Since the ODE is not exact, we will try to find an integrating factor to make it exact. Let \begin {align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial y} - \frac {\partial N}{\partial x} \right ) \\ &=\frac {1}{x}\left ( \left ( -D\left (f \right )\left (x y \right ) x y -f \left (x y \right )\right ) - \left (1 \right ) \right ) \\ &=\frac {-D\left (f \right )\left (x y \right ) x y -f \left (x y \right )-1}{x} \end {align*}
Since \(A\) depends on \(y\), it can not be used to obtain an integrating factor. We will now try a second method to find an integrating factor. Let \begin {align*} B &= \frac {1}{M} \left ( \frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} \right ) \\ &=-\frac {1}{f \left (x y \right ) y}\left ( \left ( 1\right ) - \left (-D\left (f \right )\left (x y \right ) x y -f \left (x y \right ) \right ) \right ) \\ &=\frac {-D\left (f \right )\left (x y \right ) x y -f \left (x y \right )-1}{f \left (x y \right ) y} \end {align*}
Since \(B\) depends on \(x\), it can not be used to obtain an integrating factor.We will now try a third method to find an integrating factor. Let \[ R = \frac { \frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} } {x M - y N} \] \(R\) is now checked to see if it is a function of only \(t=xy\). Therefore \begin {align*} R &= \frac { \frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} } {x M - y N} \\ &= \frac {\left (1\right )-\left (-D\left (f \right )\left (x y \right ) x y -f \left (x y \right )\right )} {x\left (-f \left (x y \right ) y\right ) - y\left (x\right )} \\ &= \frac {-D\left (f \right )\left (x y \right ) x y -f \left (x y \right )-1}{x y \left (f \left (x y \right )+1\right )} \end {align*}
Replacing all powers of terms \(xy\) by \(t\) gives \[ R = \frac {-D\left (f \right )\left (t \right ) t -f \left (t \right )-1}{t \left (f \left (t \right )+1\right )} \] Since \(R\) depends on \(t\) only, then it can be used to find an integrating factor. Let the integrating factor be \(\mu \) then \begin {align*} \mu &= e^{\int R \mathop {\mathrm {d}t}} \\ &= e^{\int \left (\frac {-D\left (f \right )\left (t \right ) t -f \left (t \right )-1}{t \left (f \left (t \right )+1\right )}\right )\mathop {\mathrm {d}t} } \end {align*}
Could not integrate the above. Unable to complete solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x -y f \left (y x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y f \left (y x \right )}{x} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying homogeneous types: trying homogeneous G <- homogeneous successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 29
dsolve(x*diff(y(x),x) - y(x)*f(x*y(x))=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\operatorname {RootOf}\left (-\ln \left (x \right )+c_{1} +\int _{}^{\textit {\_Z}}\frac {1}{\textit {\_a} \left (1+f \left (\textit {\_a} \right )\right )}d \textit {\_a} \right )}{x} \]
✓ Solution by Mathematica
Time used: 0.225 (sec). Leaf size: 115
DSolve[x*y'[x] - y[x]*f[x*y[x]]==0,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^{y(x)}\left (\frac {1}{(-f(x K[2])-1) K[2]}-\int _1^x\left (\frac {f'(K[1] K[2])}{f(K[1] K[2])+1}-\frac {f(K[1] K[2]) f'(K[1] K[2])}{(f(K[1] K[2])+1)^2}\right )dK[1]\right )dK[2]+\int _1^x\frac {f(K[1] y(x))}{(f(K[1] y(x))+1) K[1]}dK[1]=c_1,y(x)\right ] \]