34.9 problem 9

Internal problem ID [11096]
Internal file name [OUTPUT/10353_Wednesday_January_24_2024_10_18_12_PM_31832114/index.tex]

Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section: Chapter 2, Second-Order Differential Equations. section 2.1.3-1. Equations with exponential functions
Problem number: 9.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_bessel_ode_form_A", "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)]`]]

\[ \boxed {y^{\prime \prime }-a y^{\prime }+b \,{\mathrm e}^{2 a x} y=0} \]

34.9.1 Solving as second order change of variable on x method 2 ode

In normal form the ode \begin {align*} y^{\prime \prime }-a y^{\prime }+b \,{\mathrm e}^{2 a x} y&=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 )&=-a\\ q \left (x \right )&=b \,{\mathrm e}^{2 a x} \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 -a d x \right )}d x\\ &= \int e^{a x} \,dx\\ &= \int {\mathrm e}^{a x}d x\\ &= \frac {{\mathrm e}^{a x}}{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 \,{\mathrm e}^{2 a x}}{{\mathrm e}^{2 a x}}\\ &= 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}^{\frac {\sqrt {-b}\, {\mathrm e}^{a x}}{a}}+c_{2} {\mathrm e}^{-\frac {\sqrt {-b}\, {\mathrm e}^{a x}}{a}} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{\frac {\sqrt {-b}\, {\mathrm e}^{a x}}{a}}+c_{2} {\mathrm e}^{-\frac {\sqrt {-b}\, {\mathrm e}^{a x}}{a}} \\ \end{align*}

Verification of solutions

\[ y = c_{1} {\mathrm e}^{\frac {\sqrt {-b}\, {\mathrm e}^{a x}}{a}}+c_{2} {\mathrm e}^{-\frac {\sqrt {-b}\, {\mathrm e}^{a x}}{a}} \] Verified OK.

34.9.2 Solving as second order change of variable on x method 1 ode

In normal form the ode \begin {align*} y^{\prime \prime }-a y^{\prime }+b \,{\mathrm e}^{2 a x} y&=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 )&=-a\\ q \left (x \right )&=b \,{\mathrm e}^{2 a x} \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 \,{\mathrm e}^{2 a x}}}{c}\tag {6} \\ \tau '' &= \frac {b \,{\mathrm e}^{2 a x} a}{c \sqrt {b \,{\mathrm e}^{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 \,{\mathrm e}^{2 a x} a}{c \sqrt {b \,{\mathrm e}^{2 a x}}}-a\frac {\sqrt {b \,{\mathrm e}^{2 a x}}}{c}}{\left (\frac {\sqrt {b \,{\mathrm e}^{2 a x}}}{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 \,{\mathrm e}^{2 a x}}d x}{c}\\ &= \frac {\sqrt {b \,{\mathrm e}^{2 a x}}}{c a} \end {align*}

Substituting the above into the solution obtained gives \[ y = c_{1} \cos \left (\frac {\sqrt {b}\, {\mathrm e}^{a x}}{a}\right )+c_{2} \sin \left (\frac {\sqrt {b}\, {\mathrm e}^{a x}}{a}\right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \cos \left (\frac {\sqrt {b}\, {\mathrm e}^{a x}}{a}\right )+c_{2} \sin \left (\frac {\sqrt {b}\, {\mathrm e}^{a x}}{a}\right ) \\ \end{align*}

Verification of solutions

\[ y = c_{1} \cos \left (\frac {\sqrt {b}\, {\mathrm e}^{a x}}{a}\right )+c_{2} \sin \left (\frac {\sqrt {b}\, {\mathrm e}^{a x}}{a}\right ) \] Verified OK.

34.9.3 Solving as second order bessel ode form A ode

Writing the ode as \begin {align*} y^{\prime \prime }-a y^{\prime }+b \,{\mathrm e}^{2 a x} y = 0\tag {1} \end {align*}

An ode of the form\begin {equation} ay^{\prime \prime }+by^{\prime }+(ce^{rx}+m)y=0\tag {1} \end {equation} can be transformed to Bessel ode using the transformation\begin {align*} x & =\ln \left ( t\right ) \\ e^{x} & =t \end {align*}

Where \(a,b,c,m\) are not functions of \(x\) and where \(b\) and \(m\) are allowed to be be zero. Using this transformation gives\begin {align} \frac {dy}{dx} & =\frac {dy}{dt}\frac {dt}{dx}\nonumber \\ & =\frac {dy}{dt}e^{x}\nonumber \\ & =t\frac {dy}{dt}\tag {2} \end {align}

And\begin {align} \frac {d^{2}y}{dx^{2}} & =\frac {d}{dx}\left ( \frac {dy}{dx}\right ) \nonumber \\ & =\frac {d}{dx}\left ( t\frac {dy}{dt}\right ) \nonumber \\ & =\frac {d}{dt}\frac {dt}{dx}\left ( t\frac {dy}{dt}\right ) \nonumber \\ & =\frac {dt}{dx}\frac {d}{dt}\left ( t\frac {dy}{dt}\right ) \nonumber \\ & =t\frac {d}{dt}\left ( t\frac {dy}{dt}\right ) \nonumber \\ & =t\left ( \frac {dy}{dt}+t\frac {d^{2}y}{dt^{2}}\right ) \tag {3} \end {align}

Substituting (2,3) into (1) gives\begin {align} at\left ( \frac {dy}{dt}+t\frac {d^{2}y}{dt^{2}}\right ) +bt\frac {dy}{dt}+(ce^{rx}+m)y & =0\nonumber \\ \left ( aty^{\prime }+at^{2}y^{\prime \prime }\right ) +bty^{\prime }+(ct^{r}+m)y & =0\nonumber \\ at^{2}y^{\prime \prime }+\left ( b+a\right ) ty^{\prime }+(ct^{r}+m)y & =0\nonumber \\ t^{2}y^{\prime \prime }+\frac {b+a}{a}ty^{\prime }+\left ( \frac {c}{a}t^{r}+\frac {m}{a}\right ) y & =0\tag {4} \end {align}

Which is Bessel ODE. Comparing the above to the general known Bowman form of Bessel ode which is\begin {equation} t^{2}y^{\prime \prime }+\left ( 1-2\alpha \right ) ty^{\prime }+\left ( \beta ^{2}\gamma ^{2}t^{2\gamma }-\left ( n^{2}\gamma ^{2}-\alpha ^{2}\right ) \right ) y=0\tag {C} \end {equation} And now comparing (4) and (C) shows that\begin {align} \left ( 1-2\alpha \right ) & =\frac {b+a}{a}\tag {5}\\ \beta ^{2}\gamma ^{2} & =\frac {c}{a}\tag {6}\\ 2\gamma & =r\tag {7}\\ \left ( n^{2}\gamma ^{2}-\alpha ^{2}\right ) & =-\frac {m}{a}\tag {8} \end {align}

(5) gives \(\alpha =\frac {1}{2}-\frac {b+a}{2a}\). (7) gives \(\gamma =\frac {r}{2}\). (8) now becomes \(\left ( n^{2}\left ( \frac {r}{2}\right ) ^{2}-\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}\right ) =-\frac {m}{a}\) or \(n^{2}=\frac {-\frac {m}{a}+\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}}{\left ( \frac {r}{2}\right ) ^{2}}\). Hence \(n=\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}}\ \)by taking the positive root. And finally (6) gives \(\beta ^{2}=\frac {c}{a\gamma ^{2}}\) or \(\beta =\sqrt {\frac {c}{a}}\frac {1}{\gamma }=\sqrt {\frac {c}{a}}\frac {2}{r}\) (also taking the positive root). Hence\begin {align*} \alpha & =\frac {1}{2}-\frac {b+a}{2a}\\ n & =\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}}\\ \beta & =\sqrt {\frac {c}{a}}\frac {2}{r}\\ \gamma & =\frac {r}{2} \end {align*}

But the solution to (C) which is general form of Bessel ode is known and given by \[ y\left ( t\right ) =t^{\alpha }\left ( c_{1}J_{n}\left ( \beta t^{\gamma }\right ) +c_{2}Y_{n}\left ( \beta t^{\gamma }\right ) \right ) \] Substituting the above values found into this solution gives\[ y\left ( t\right ) =t^{\frac {1}{2}-\frac {b+a}{2a}}\left ( c_{1}J_{\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}t^{\frac {r}{2}}\right ) +c_{2}Y_{\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}t^{\frac {r}{2}}\right ) \right ) \] Since \(e^{x}=t\) then the above becomes\begin {align} y\left ( x\right ) & =e^{x\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) }\left ( c_{1}J_{\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) +c_{2}Y_{\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) \right ) \nonumber \\ & =e^{x\left ( \frac {-b}{2a}\right ) }\left ( c_{1}J_{\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {-b}{2a}\right ) ^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) +c_{2}Y_{\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {-b}{2a}\right ) ^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) \right ) \nonumber \\ & =e^{x\left ( \frac {-b}{2a}\right ) }\left ( c_{1}J_{\frac {2}{r}\sqrt {-\frac {m}{a}+\frac {b^{2}}{4a^{2}}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) +c_{2}Y_{\frac {2}{r}\sqrt {-\frac {m}{a}+\frac {b^{2}}{4a^{2}}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) \right ) \nonumber \\ & =e^{x\left ( \frac {-b}{2a}\right ) }\left ( c_{1}J_{\frac {2}{r}\sqrt {-\frac {4ma+b^{2}}{4a^{2}}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) +c_{2}Y_{\frac {2}{r}\sqrt {-\frac {4ma+b^{2}}{4a^{2}}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) \right ) \nonumber \\ & =e^{x\left ( \frac {-b}{2a}\right ) }\left ( c_{1}J_{\frac {1}{ra}\sqrt {-4ma+b^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) +c_{2}Y_{\frac {1}{ra}\sqrt {-4ma+b^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) \right ) \tag {9} \end {align}

Equation (9) above is the solution to \(ay^{\prime \prime }+by^{\prime }+(ce^{rx}+m)y=0\). Therefore we just need now to compare this form to the ode given and use (9) to obtain the final solution.

Comparing form (1) to the ode we are solving shows that \begin {align*} a &= 1\\ b &= -a\\ c &= b\\ r &= 2 a\\ m &= 0 \end {align*}

Substituting these in (9) gives the solution as \begin {align*} y&=\frac {c_{1} {\mathrm e}^{\frac {a x}{2}} \sqrt {2}\, \sin \left (\frac {\sqrt {b}\, {\mathrm e}^{a x}}{a}\right )}{\sqrt {\pi }\, \sqrt {\frac {\sqrt {b}\, {\mathrm e}^{a x}}{a}}}-\frac {c_{2} {\mathrm e}^{\frac {a x}{2}} \sqrt {2}\, \cos \left (\frac {\sqrt {b}\, {\mathrm e}^{a x}}{a}\right )}{\sqrt {\pi }\, \sqrt {\frac {\sqrt {b}\, {\mathrm e}^{a x}}{a}}} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} {\mathrm e}^{\frac {a x}{2}} \sqrt {2}\, \sin \left (\frac {\sqrt {b}\, {\mathrm e}^{a x}}{a}\right )}{\sqrt {\pi }\, \sqrt {\frac {\sqrt {b}\, {\mathrm e}^{a x}}{a}}}-\frac {c_{2} {\mathrm e}^{\frac {a x}{2}} \sqrt {2}\, \cos \left (\frac {\sqrt {b}\, {\mathrm e}^{a x}}{a}\right )}{\sqrt {\pi }\, \sqrt {\frac {\sqrt {b}\, {\mathrm e}^{a x}}{a}}} \\ \end{align*}

Verification of solutions

\[ y = \frac {c_{1} {\mathrm e}^{\frac {a x}{2}} \sqrt {2}\, \sin \left (\frac {\sqrt {b}\, {\mathrm e}^{a x}}{a}\right )}{\sqrt {\pi }\, \sqrt {\frac {\sqrt {b}\, {\mathrm e}^{a x}}{a}}}-\frac {c_{2} {\mathrm e}^{\frac {a x}{2}} \sqrt {2}\, \cos \left (\frac {\sqrt {b}\, {\mathrm e}^{a x}}{a}\right )}{\sqrt {\pi }\, \sqrt {\frac {\sqrt {b}\, {\mathrm e}^{a x}}{a}}} \] 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: 0.016 (sec). Leaf size: 33

dsolve(diff(y(x),x$2)-a*diff(y(x),x)+b*exp(2*a*x)*y(x)=0,y(x), singsol=all)
 

\[ y \left (x \right ) = c_{1} \sin \left (\frac {\sqrt {b}\, {\mathrm e}^{a x}}{a}\right )+c_{2} \cos \left (\frac {\sqrt {b}\, {\mathrm e}^{a x}}{a}\right ) \]

✓ Solution by Mathematica

Time used: 0.061 (sec). Leaf size: 42

DSolve[y''[x]-a*y'[x]+b*Exp[2*a*x]*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to c_1 \cos \left (\frac {\sqrt {b} e^{a x}}{a}\right )+c_2 \sin \left (\frac {\sqrt {b} e^{a x}}{a}\right ) \]