Internal problem ID [6853]
Internal file name [OUTPUT/6100_Friday_July_29_2022_03_09_14_AM_38707999/index.tex]
Book: Elementary differential equations. By Earl D. Rainville, Phillip E. Bedient. Macmilliam
Publishing Co. NY. 6th edition. 1981.
Section: CHAPTER 16. Nonlinear equations. Section 101. Independent variable missing.
EXERCISES Page 324
Problem number: 36.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_y", "second_order_nonlinear_solved_by_mainardi_lioville_method"
Maple gives the following as the ode type
[[_2nd_order, _missing_y], _Liouville, [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {x y^{\prime \prime }-y^{\prime } \left (2-3 y^{\prime } x \right )=0} \]
This is second order ode with missing dependent variable \(y\). Let \begin {align*} p(x) &= y^{\prime } \end {align*}
Then \begin {align*} p'(x) &= y^{\prime \prime } \end {align*}
Hence the ode becomes \begin {align*} x p^{\prime }\left (x \right )+\left (3 p \left (x \right ) x -2\right ) p \left (x \right ) = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. Writing the ode as \begin {align*} p^{\prime }\left (x \right )&=-\frac {\left (3 p x -2\right ) p}{x}\\ p^{\prime }\left (x \right )&= \omega \left ( x,p\right ) \end {align*}
The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{x}+\omega \left ( \eta _{p}-\xi _{x}\right ) -\omega ^{2}\xi _{p}-\omega _{x}\xi -\omega _{p}\eta =0\tag {A} \end {align*}
The type of this ode is known. It is of type Bernoulli. Therefore we do not need
to solve the PDE (A), and can just use the lookup table shown below to find \(\xi ,\eta \)
|
ODE class |
Form |
\(\xi \) |
\(\eta \) |
|
linear ode |
\(y'=f(x) y(x) +g(x)\) |
\(0\) |
\(e^{\int fdx}\) |
|
separable ode |
\(y^{\prime }=f\left ( x\right ) g\left ( y\right ) \) |
\(\frac {1}{f}\) |
\(0\) |
|
quadrature ode |
\(y^{\prime }=f\left ( x\right ) \) |
\(0\) |
\(1\) |
|
quadrature ode |
\(y^{\prime }=g\left ( y\right ) \) |
\(1\) |
\(0\) |
|
homogeneous ODEs of Class A |
\(y^{\prime }=f\left ( \frac {y}{x}\right ) \) |
\(x\) |
\(y\) |
|
homogeneous ODEs of Class C |
\(y^{\prime }=\left ( a+bx+cy\right ) ^{\frac {n}{m}}\) |
\(1\) |
\(-\frac {b}{c}\) |
|
homogeneous class D |
\(y^{\prime }=\frac {y}{x}+g\left ( x\right ) F\left (\frac {y}{x}\right ) \) |
\(x^{2}\) |
\(xy\) |
|
First order special form ID 1 |
\(y^{\prime }=g\left ( x\right ) e^{h\left (x\right ) +by}+f\left ( x\right ) \) |
\(\frac {e^{-\int bf\left ( x\right )dx-h\left ( x\right ) }}{g\left ( x\right ) }\) |
\(\frac {f\left ( x\right )e^{-\int bf\left ( x\right ) dx-h\left ( x\right ) }}{g\left ( x\right ) }\) |
|
polynomial type ode |
\(y^{\prime }=\frac {a_{1}x+b_{1}y+c_{1}}{a_{2}x+b_{2}y+c_{2}}\) |
\(\frac {a_{1}b_{2}x-a_{2}b_{1}x-b_{1}c_{2}+b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}\) |
\(\frac {a_{1}b_{2}y-a_{2}b_{1}y-a_{1}c_{2}-a_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}\) |
|
Bernoulli ode |
\(y^{\prime }=f\left ( x\right ) y+g\left ( x\right ) y^{n}\) |
\(0\) |
\(e^{-\int \left ( n-1\right ) f\left ( x\right ) dx}y^{n}\) |
|
Reduced Riccati |
\(y^{\prime }=f_{1}\left ( x\right ) y+f_{2}\left ( x\right )y^{2}\) |
\(0\) |
\(e^{-\int f_{1}dx}\) |
|
|
|||
|
|
|||
The above table shows that \begin {align*} \xi \left (x,p\right ) &=0\\ \tag {A1} \eta \left (x,p\right ) &=\frac {p^{2}}{x^{2}} \end {align*}
The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,p\right ) \to \left ( R,S \right )\) 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 p}{\eta } = dS \tag {1} \end {align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial p}\right ) S(x,p) = 1\). 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 \(S(R)\). Since \(\xi =0\) then in this special case \begin {align*} R = x \end {align*}
\(S\) is found from \begin {align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{\frac {p^{2}}{x^{2}}}} dy \end {align*}
Which results in \begin {align*} S&= -\frac {x^{2}}{p} \end {align*}
Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by evaluating \begin {align*} \frac {dS}{dR} &= \frac { S_{x} + \omega (x,p) S_{p} }{ R_{x} + \omega (x,p) R_{p} }\tag {2} \end {align*}
Where in the above \(R_{x},R_{p},S_{x},S_{p}\) are all partial derivatives and \(\omega (x,p)\) is the right hand side of the original ode given by \begin {align*} \omega (x,p) &= -\frac {\left (3 p x -2\right ) p}{x} \end {align*}
Evaluating all the partial derivatives gives \begin {align*} R_{x} &= 1\\ R_{p} &= 0\\ S_{x} &= -\frac {2 x}{p}\\ S_{p} &= \frac {x^{2}}{p^{2}} \end {align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates. \begin {align*} \frac {dS}{dR} &= -3 x^{2}\tag {2A} \end {align*}
We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,p\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives \begin {align*} \frac {dS}{dR} &= -3 R^{2} \end {align*}
The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates \(R,S\). Integrating the above gives \begin {align*} S \left (R \right ) = -R^{3}+c_{1}\tag {4} \end {align*}
To complete the solution, we just need to transform (4) back to \(x,p\) coordinates. This results in \begin {align*} -\frac {x^{2}}{p \left (x \right )} = -x^{3}+c_{1} \end {align*}
Which simplifies to \begin {align*} -\frac {x^{2}}{p \left (x \right )} = -x^{3}+c_{1} \end {align*}
Which gives \begin {align*} p \left (x \right ) = -\frac {x^{2}}{-x^{3}+c_{1}} \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = -\frac {x^{2}}{-x^{3}+c_{1}} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -\frac {x^{2}}{-x^{3}+c_{1}}\,\mathop {\mathrm {d}x}}\\ &= \frac {\ln \left (x^{3}-c_{1} \right )}{3}+c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\ln \left (x^{3}-c_{1} \right )}{3}+c_{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {\ln \left (x^{3}-c_{1} \right )}{3}+c_{2} \] Verified OK.
The ode has the Liouville form given by \begin {align*} y^{\prime \prime }+ f(x) y^{\prime } + g(y) {y^{\prime }}^{2} &= 0 \tag {1A} \end {align*}
Where in this problem \begin {align*} f(x) &= -\frac {2}{x}\\ g(y) &= 3 \end {align*}
Dividing through by \(y^{\prime }\) then Eq (1A) becomes \begin {align*} \frac {y^{\prime \prime }}{y^{\prime }}+ f + g y^{\prime } &= 0 \tag {2A} \end {align*}
But the first term in Eq (2A) can be written as \begin {align*} \frac {y^{\prime \prime }}{y^{\prime }}&= \frac {d}{dx} \ln \left ( y^{\prime } \right )\tag {3A} \end {align*}
And the last term in Eq (2A) can be written as \begin {align*} g \frac {dy}{dx}&= \left ( \frac {d}{dy} \int g d y\right ) \frac {dy}{dx} \\ &= \frac {d}{dx} \int g d y\tag {4A} \end {align*}
Substituting (3A,4A) back into (2A) gives \begin {align*} \frac {d}{dx} \ln \left ( y^{\prime } \right ) + \frac {d}{dx} \int g d y &= -f \tag {5A} \end {align*}
Integrating the above w.r.t. \(x\) gives \begin {align*} \ln \left ( y^{\prime } \right ) + \int g d y &= - \int f d x + c_{1} \end {align*}
Where \(c_1\) is arbitrary constant. Taking the exponential of the above gives \begin {align*} y^{\prime } &= c_{2} e^{\int -g d y}\, e^{\int -f d x}\tag {6A} \end {align*}
Where \(c_{2}\) is a new arbitrary constant. But since \(g=3\) and \(f=-\frac {2}{x}\), then \begin {align*} \int -g d y &= \int \left (-3\right )d y\\ &= -3 y\\ \int -f d x &= \int \frac {2}{x}d x\\ &= 2 \ln \left (x \right ) \end {align*}
Substituting the above into Eq(6A) gives \[ y^{\prime } = c_{2} {\mathrm e}^{-3 y} x^{2} \] Which is now solved as first order separable ode. In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= c_{2} {\mathrm e}^{-3 y} x^{2} \end {align*}
Where \(f(x)=c_{2} x^{2}\) and \(g(y)={\mathrm e}^{-3 y}\). Integrating both sides gives \begin{align*} \frac {1}{{\mathrm e}^{-3 y}} \,dy &= c_{2} x^{2} \,d x \\ \int { \frac {1}{{\mathrm e}^{-3 y}} \,dy} &= \int {c_{2} x^{2} \,d x} \\ \frac {{\mathrm e}^{3 y}}{3}&=\frac {c_{2} x^{3}}{3}+c_{3} \\ \end{align*} The solution is \[ \frac {{\mathrm e}^{3 y}}{3}-\frac {c_{2} x^{3}}{3}-c_{3} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} \frac {{\mathrm e}^{3 y}}{3}-\frac {c_{2} x^{3}}{3}-c_{3} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {{\mathrm e}^{3 y}}{3}-\frac {c_{2} x^{3}}{3}-c_{3} = 0 \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville <- 2nd_order Liouville successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 16
dsolve(x*diff(y(x),x$2)=diff(y(x),x)*(2-3*x*diff(y(x),x)),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\ln \left (c_{1} x^{3}+3 c_{2} \right )}{3} \]
✓ Solution by Mathematica
Time used: 0.267 (sec). Leaf size: 19
DSolve[x*y''[x]==y'[x]*(2-3*x*y'[x]),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{3} \log \left (x^3+c_1\right )+c_2 \]