Internal problem ID [9736]
Internal file name [OUTPUT/8678_Monday_June_06_2022_05_09_43_AM_46421080/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1409.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_change_of_variable_on_x_method_1", "second_order_change_of_variable_on_x_method_2"
Maple gives the following as the ode type
[[_Emden, _Fowler], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]
\[ \boxed {y^{\prime \prime }+a \,x^{2 a -1} x^{-2 a} y^{\prime }+b^{2} x^{-2 a} y=0} \]
In normal form the ode \begin {align*} a \,x^{2 a -1} y^{\prime }+b^{2} y+y^{\prime \prime } x^{2 a}&=0 \tag {1} \end {align*}
Becomes \begin {align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end {align*}
Where \begin {align*} p \left (x \right )&=\frac {a}{x}\\ q \left (x \right )&=b^{2} x^{-2 a} \end {align*}
Applying change of variables \(\tau = g \left (x \right )\) to (2) gives \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end {align*}
Where \(\tau \) is the new independent variable, and \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end {align*}
Let \(p_{1} = 0\). Eq (4) simplifies to \begin {align*} \tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )&=0 \end {align*}
This ode is solved resulting in \begin {align*} \tau &= \int {\mathrm e}^{-\left (\int p \left (x \right )d x \right )}d x\\ &= \int {\mathrm e}^{-\left (\int \frac {a}{x}d x \right )}d x\\ &= \int e^{-a \ln \left (x \right )} \,dx\\ &= \int x^{-a}d x\\ &= -\frac {x^{1-a}}{-1+a}\tag {6} \end {align*}
Using (6) to evaluate \(q_{1}\) from (5) gives \begin {align*} q_{1} \left (\tau \right ) &= \frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &= \frac {b^{2} x^{-2 a}}{x^{-2 a}}\\ &= b^{2}\tag {7} \end {align*}
Substituting the above in (3) and noting that now \(p_{1} = 0\) results in \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+q_{1} y \left (\tau \right )&=0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+b^{2} y \left (\tau \right )&=0 \end {align*}
The above ode is now solved for \(y \left (\tau \right )\).This is second order with constant coefficients homogeneous ODE. In standard form the ODE is \[ A y''(\tau ) + B y'(\tau ) + C y(\tau ) = 0 \] Where in the above \(A=1, B=0, C=b^{2}\). Let the solution be \(y \left (\tau \right )=e^{\lambda \tau }\). Substituting this into the ODE gives \[ \lambda ^{2} {\mathrm e}^{\lambda \tau }+b^{2} {\mathrm e}^{\lambda \tau } = 0 \tag {1} \] Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda \tau }\) gives \[ b^{2}+\lambda ^{2} = 0 \tag {2} \] Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula \[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \] Substituting \(A=1, B=0, C=b^{2}\) into the above gives \begin {align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {0^2 - (4) \left (1\right )\left (b^{2}\right )}\\ &= \pm \sqrt {-b^{2}} \end {align*}
Hence \begin{align*} \lambda _1 &= + \sqrt {-b^{2}} \\ \lambda _2 &= - \sqrt {-b^{2}} \\ \end{align*} Which simplifies to \begin{align*} \lambda _1 &= \sqrt {-b^{2}} \\ \lambda _2 &= -\sqrt {-b^{2}} \\ \end{align*} Since roots are real and distinct, then the solution is \begin{align*} y \left (\tau \right ) &= c_{1} e^{\lambda _1 \tau } + c_{2} e^{\lambda _2 \tau } \\ y \left (\tau \right ) &= c_{1} e^{\left (\sqrt {-b^{2}}\right )\tau } +c_{2} e^{\left (-\sqrt {-b^{2}}\right )\tau } \\ \end{align*} Or \[ y \left (\tau \right ) =c_{1} {\mathrm e}^{\sqrt {-b^{2}}\, \tau }+c_{2} {\mathrm e}^{-\sqrt {-b^{2}}\, \tau } \] The above solution is now transformed back to \(y\) using (6) which results in \begin {align*} y &= c_{1} {\mathrm e}^{-\frac {\sqrt {-b^{2}}\, x^{1-a}}{-1+a}}+c_{2} {\mathrm e}^{\frac {\sqrt {-b^{2}}\, x^{1-a}}{-1+a}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-\frac {\sqrt {-b^{2}}\, x^{1-a}}{-1+a}}+c_{2} {\mathrm e}^{\frac {\sqrt {-b^{2}}\, x^{1-a}}{-1+a}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{-\frac {\sqrt {-b^{2}}\, x^{1-a}}{-1+a}}+c_{2} {\mathrm e}^{\frac {\sqrt {-b^{2}}\, x^{1-a}}{-1+a}} \] Verified OK.
In normal form the ode \begin {align*} a \,x^{2 a -1} y^{\prime }+b^{2} y+y^{\prime \prime } x^{2 a}&=0 \tag {1} \end {align*}
Becomes \begin {align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end {align*}
Where \begin {align*} p \left (x \right )&=\frac {a}{x}\\ q \left (x \right )&=b^{2} x^{-2 a} \end {align*}
Applying change of variables \(\tau = g \left (x \right )\) to (2) results \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end {align*}
Where \(\tau \) is the new independent variable, and \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end {align*}
Let \(q_1=c^2\) where \(c\) is some constant. Therefore from (5) \begin {align*} \tau ' &= \frac {1}{c}\sqrt {q}\\ &= \frac {\sqrt {b^{2} x^{-2 a}}}{c}\tag {6} \\ \tau '' &= -\frac {b^{2} x^{-2 a} a}{c \sqrt {b^{2} x^{-2 a}}\, x} \end {align*}
Substituting the above into (4) results in \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &=\frac {-\frac {b^{2} x^{-2 a} a}{c \sqrt {b^{2} x^{-2 a}}\, x}+\frac {a}{x}\frac {\sqrt {b^{2} x^{-2 a}}}{c}}{\left (\frac {\sqrt {b^{2} x^{-2 a}}}{c}\right )^2} \\ &=0 \end {align*}
Therefore ode (3) now becomes \begin {align*} y \left (\tau \right )'' + p_1 y \left (\tau \right )' + q_1 y \left (\tau \right ) &= 0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+c^{2} y \left (\tau \right ) &= 0 \tag {7} \end {align*}
The above ode is now solved for \(y \left (\tau \right )\). Since the ode is now constant coefficients, it can be easily solved to give \begin {align*} y \left (\tau \right ) &= c_{1} \cos \left (c \tau \right )+c_{2} \sin \left (c \tau \right ) \end {align*}
Now from (6) \begin {align*} \tau &= \int \frac {1}{c} \sqrt q \,dx \\ &= \frac {\int \sqrt {b^{2} x^{-2 a}}d x}{c}\\ &= -\frac {x \sqrt {b^{2} x^{-2 a}}}{c \left (-1+a \right )} \end {align*}
Substituting the above into the solution obtained gives \[ y = c_{1} \cos \left (\frac {b \,x^{1-a}}{-1+a}\right )-c_{2} \sin \left (\frac {b \,x^{1-a}}{-1+a}\right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \cos \left (\frac {b \,x^{1-a}}{-1+a}\right )-c_{2} \sin \left (\frac {b \,x^{1-a}}{-1+a}\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \cos \left (\frac {b \,x^{1-a}}{-1+a}\right )-c_{2} \sin \left (\frac {b \,x^{1-a}}{-1+a}\right ) \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] <- linear_1 successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 39
dsolve(diff(diff(y(x),x),x) = -a*x^(2*a-1)/(x^(2*a))*diff(y(x),x)-b^2/(x^(2*a))*y(x),y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \sin \left (\frac {b \,x^{-a +1}}{a -1}\right )+c_{2} \cos \left (\frac {b \,x^{-a +1}}{a -1}\right ) \]
✓ Solution by Mathematica
Time used: 0.044 (sec). Leaf size: 44
DSolve[y''[x] == -((b^2*y[x])/x^(2*a)) - (a*y'[x])/x,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 \cos \left (\frac {b x^{1-a}}{a-1}\right )+c_2 \sin \left (\frac {b x^{1-a}}{1-a}\right ) \]