Internal problem ID [9410]
Internal file name [OUTPUT/8348_Monday_June_06_2022_02_48_33_AM_87461315/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1079.
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
[[_2nd_order, _with_linear_symmetries], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]
\[ \boxed {y^{\prime \prime }-\frac {a f^{\prime }\left (x \right ) y^{\prime }}{f \left (x \right )}+b f \left (x \right )^{2 a} y=0} \]
In normal form the ode \begin {align*} b f \left (x \right )^{2 a +1} y-a f^{\prime }\left (x \right ) y^{\prime }+y^{\prime \prime } f \left (x \right )&=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 f^{\prime }\left (x \right )}{f \left (x \right )}\\ q \left (x \right )&=b f \left (x \right )^{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 f^{\prime }\left (x \right )}{f \left (x \right )}d x \right )}d x\\ &= \int e^{a \ln \left (f \left (x \right )\right )} \,dx\\ &= \int f \left (x \right )^{a}d x\\ &= \int f \left (x \right )^{a}d x\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 f \left (x \right )^{2 a}}{f \left (x \right )^{2 a}}\\ &= b\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 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 \(A=1, B=0, C=b\). Let the solution be \(y \left (\tau \right )=e^{\lambda \tau }\). Substituting this into the ODE gives \[ \lambda ^{2} {\mathrm e}^{\lambda \tau }+b \,{\mathrm e}^{\lambda \tau } = 0 \tag {1} \] Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda \tau }\) gives \[ \lambda ^{2}+b = 0 \tag {2} \] 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 \(A=1, B=0, C=b\) 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\right )}\\ &= \pm \sqrt {-b} \end {align*}
Hence \begin{align*} \lambda _1 &= + \sqrt {-b} \\ \lambda _2 &= - \sqrt {-b} \\ \end{align*} Which simplifies to \begin{align*} \lambda _1 &= \sqrt {-b} \\ \lambda _2 &= -\sqrt {-b} \\ \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}\right )\tau } +c_{2} e^{\left (-\sqrt {-b}\right )\tau } \\ \end{align*} Or \[ y \left (\tau \right ) =c_{1} {\mathrm e}^{\sqrt {-b}\, \tau }+c_{2} {\mathrm e}^{-\sqrt {-b}\, \tau } \] The above solution is now transformed back to \(y\) using (6) which results in \begin {align*} y &= c_{1} {\mathrm e}^{\sqrt {-b}\, \left (\int f \left (x \right )^{a}d x \right )}+c_{2} {\mathrm e}^{-\sqrt {-b}\, \left (\int f \left (x \right )^{a}d x \right )} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{\sqrt {-b}\, \left (\int f \left (x \right )^{a}d x \right )}+c_{2} {\mathrm e}^{-\sqrt {-b}\, \left (\int f \left (x \right )^{a}d x \right )} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{\sqrt {-b}\, \left (\int f \left (x \right )^{a}d x \right )}+c_{2} {\mathrm e}^{-\sqrt {-b}\, \left (\int f \left (x \right )^{a}d x \right )} \] Verified OK.
In normal form the ode \begin {align*} b f \left (x \right )^{2 a +1} y-a f^{\prime }\left (x \right ) y^{\prime }+y^{\prime \prime } f \left (x \right )&=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 f^{\prime }\left (x \right )}{f \left (x \right )}\\ q \left (x \right )&=b f \left (x \right )^{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 f \left (x \right )^{2 a}}}{c}\tag {6} \\ \tau '' &= \frac {b f \left (x \right )^{2 a} a f^{\prime }\left (x \right )}{c \sqrt {b f \left (x \right )^{2 a}}\, f \left (x \right )} \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 f \left (x \right )^{2 a} a f^{\prime }\left (x \right )}{c \sqrt {b f \left (x \right )^{2 a}}\, f \left (x \right )}-\frac {a f^{\prime }\left (x \right )}{f \left (x \right )}\frac {\sqrt {b f \left (x \right )^{2 a}}}{c}}{\left (\frac {\sqrt {b f \left (x \right )^{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 f \left (x \right )^{2 a}}d x}{c}\\ &= \frac {\int \sqrt {b f \left (x \right )^{2 a}}d x}{c} \end {align*}
Substituting the above into the solution obtained gives \[ y = c_{1} \cos \left (\sqrt {b}\, \left (\int \sqrt {f \left (x \right )^{2 a}}d x \right )\right )+c_{2} \sin \left (\sqrt {b}\, \left (\int \sqrt {f \left (x \right )^{2 a}}d x \right )\right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \cos \left (\sqrt {b}\, \left (\int \sqrt {f \left (x \right )^{2 a}}d x \right )\right )+c_{2} \sin \left (\sqrt {b}\, \left (\int \sqrt {f \left (x \right )^{2 a}}d x \right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \cos \left (\sqrt {b}\, \left (\int \sqrt {f \left (x \right )^{2 a}}d x \right )\right )+c_{2} \sin \left (\sqrt {b}\, \left (\int \sqrt {f \left (x \right )^{2 a}}d x \right )\right ) \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] Solution is available but has integrals. Trying a simpler solution using Kovacics algorithm... Solution via Kovacic is not simpler. Returning default solution <- linear_1 successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 35
dsolve(diff(diff(y(x),x),x)-a*diff(f(x),x)/f(x)*diff(y(x),x)+b*f(x)^(2*a)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} {\mathrm e}^{i \sqrt {b}\, \left (\int f \left (x \right )^{a}d x \right )}+c_{2} {\mathrm e}^{-i \sqrt {b}\, \left (\int f \left (x \right )^{a}d x \right )} \]
✓ Solution by Mathematica
Time used: 0.556 (sec). Leaf size: 307
DSolve[b*f[x]^(2*a)*y[x] - (a*Derivative[1][f][x]*y'[x])/f[x] + y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {\sqrt {c_1} \exp \left (-\int _1^x-i \sqrt {b} f(K[1])^adK[1]-c_2\right ) \left (-1+\exp \left (2 \left (\int _1^x-i \sqrt {b} f(K[1])^adK[1]+c_2\right )\right )\right )}{\sqrt {2}} \\ y(x)\to \frac {\sqrt {c_1} \exp \left (-\int _1^x-i \sqrt {b} f(K[1])^adK[1]-c_2\right ) \left (-1+\exp \left (2 \left (\int _1^x-i \sqrt {b} f(K[1])^adK[1]+c_2\right )\right )\right )}{\sqrt {2}} \\ y(x)\to -\frac {\sqrt {c_1} \exp \left (-\int _1^xi \sqrt {b} f(K[2])^adK[2]-c_2\right ) \left (-1+\exp \left (2 \left (\int _1^xi \sqrt {b} f(K[2])^adK[2]+c_2\right )\right )\right )}{\sqrt {2}} \\ y(x)\to \frac {\sqrt {c_1} \exp \left (-\int _1^xi \sqrt {b} f(K[2])^adK[2]-c_2\right ) \left (-1+\exp \left (2 \left (\int _1^xi \sqrt {b} f(K[2])^adK[2]+c_2\right )\right )\right )}{\sqrt {2}} \\ \end{align*}