Internal problem ID [8395]
Internal file name [OUTPUT/7328_Sunday_June_05_2022_05_50_37_PM_90653100/index.tex
]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 58.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[[_homogeneous, `class G`], _Chini]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime }-a \sqrt {y}=x b} \]
Writing the ode as \begin {align*} y^{\prime }&=a \sqrt {y}+x b\\ 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}+\left (a \sqrt {y}+x b \right ) \left (b_{3}-a_{2}\right )-\left (a \sqrt {y}+x b \right )^{2} a_{3}-b \left (x a_{2}+y a_{3}+a_{1}\right )-\frac {a \left (x b_{2}+y b_{3}+b_{1}\right )}{2 \sqrt {y}} = 0 \end{equation} Putting the above in normal form gives \[ -\frac {2 y^{{3}/{2}} a^{2} a_{3}+4 y a b x a_{3}+2 \sqrt {y}\, b^{2} x^{2} a_{3}+2 y a a_{2}-a y b_{3}+4 \sqrt {y}\, b x a_{2}-2 \sqrt {y}\, b x b_{3}+2 y^{{3}/{2}} b a_{3}+2 \sqrt {y}\, b a_{1}+a x b_{2}-2 b_{2} \sqrt {y}+a b_{1}}{2 \sqrt {y}} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} -2 y^{{3}/{2}} a^{2} a_{3}-2 \sqrt {y}\, b^{2} x^{2} a_{3}-2 y^{{3}/{2}} b a_{3}-4 y a b x a_{3}-4 \sqrt {y}\, b x a_{2}+2 \sqrt {y}\, b x b_{3}-2 \sqrt {y}\, b a_{1}-a x b_{2}-2 y a a_{2}+a y b_{3}+2 b_{2} \sqrt {y}-a b_{1} = 0 \end{equation} Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \left \{x, y, \sqrt {y}, y^{{3}/{2}}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, \sqrt {y} = v_{3}, y^{{3}/{2}} = v_{4}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} -2 v_{3} b^{2} v_{1}^{2} a_{3}-4 v_{2} a b v_{1} a_{3}-2 v_{4} a^{2} a_{3}-4 v_{3} b v_{1} a_{2}+2 v_{3} b v_{1} b_{3}-2 v_{2} a a_{2}-a v_{1} b_{2}+a v_{2} b_{3}-2 v_{3} b a_{1}-2 v_{4} b a_{3}-a b_{1}+2 b_{2} v_{3} = 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} -2 v_{3} b^{2} v_{1}^{2} a_{3}-4 v_{2} a b v_{1} a_{3}+\left (-4 b a_{2}+2 b b_{3}\right ) v_{1} v_{3}-a v_{1} b_{2}+\left (-2 a a_{2}+a b_{3}\right ) v_{2}+\left (-2 b a_{1}+2 b_{2}\right ) v_{3}+\left (-2 a^{2} a_{3}-2 b a_{3}\right ) v_{4}-a b_{1} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -a b_{1}&=0\\ -a b_{2}&=0\\ -2 b^{2} a_{3}&=0\\ -4 a b a_{3}&=0\\ -2 a^{2} a_{3}-2 b a_{3}&=0\\ -2 a a_{2}+a b_{3}&=0\\ -2 b a_{1}+2 b_{2}&=0\\ -4 b a_{2}+2 b b_{3}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=a_{2}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=2 a_{2} \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 &= 2 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
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-a \sqrt {y}=x b \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=a \sqrt {y}+x b \end {array} \]
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 G trying an integrating factor from the invariance group <- integrating factor successful <- homogeneous successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 68
dsolve(diff(y(x),x) - a*sqrt(y(x)) - b*x=0,y(x), singsol=all)
\[ -\frac {\ln \left (\sqrt {y \left (x \right )}\, a x +b \,x^{2}-2 y \left (x \right )\right )}{2}+\frac {a \sqrt {y \left (x \right )}\, \operatorname {arctanh}\left (\frac {a \sqrt {y \left (x \right )}+2 b x}{\sqrt {y \left (x \right ) \left (a^{2}+8 b \right )}}\right )}{\sqrt {y \left (x \right ) \left (a^{2}+8 b \right )}}+c_{1} = 0 \]
✓ Solution by Mathematica
Time used: 0.273 (sec). Leaf size: 119
DSolve[y'[x] - a*Sqrt[y[x]] - b*x==0,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\frac {a^2 \left (-\frac {2 a \text {arctanh}\left (\frac {a^2-4 b \sqrt {\frac {a^2 y(x)}{b^2 x^2}}}{a \sqrt {a^2+8 b}}\right )}{\sqrt {a^2+8 b}}-\log \left (a^2 \left (\sqrt {\frac {a^2 y(x)}{b^2 x^2}}+1\right )-\frac {2 a^2 y(x)}{b x^2}\right )\right )}{2 b}=\frac {a^2 \log (x)}{b}+c_1,y(x)\right ] \]