Internal problem ID [11111]
Internal file name [OUTPUT/10368_Wednesday_January_24_2024_10_18_18_PM_3875151/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: 24.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_change_of_variable_on_y_method_1"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime }+\left (2 a \,{\mathrm e}^{\lambda x}+b \right ) y^{\prime }+\left (a^{2} {\mathrm e}^{2 \lambda x}+a \left (b +\lambda \right ) {\mathrm e}^{\lambda x}+c \right ) y=0} \]
In normal form the given ode is written as \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 )&=2 a \,{\mathrm e}^{\lambda x}+b\\ q \left (x \right )&={\mathrm e}^{\lambda x} a b +{\mathrm e}^{\lambda x} a \lambda +a^{2} {\mathrm e}^{2 \lambda x}+c \end {align*}
Calculating the Liouville ode invariant \(Q\) given by \begin {align*} Q &= q - \frac {p'}{2}- \frac {p^2}{4} \\ &= {\mathrm e}^{\lambda x} a b +{\mathrm e}^{\lambda x} a \lambda +a^{2} {\mathrm e}^{2 \lambda x}+c - \frac {\left (2 a \,{\mathrm e}^{\lambda x}+b\right )'}{2}- \frac {\left (2 a \,{\mathrm e}^{\lambda x}+b\right )^2}{4} \\ &= {\mathrm e}^{\lambda x} a b +{\mathrm e}^{\lambda x} a \lambda +a^{2} {\mathrm e}^{2 \lambda x}+c - \frac {\left (2 \,{\mathrm e}^{\lambda x} a \lambda \right )}{2}- \frac {\left (\left (2 a \,{\mathrm e}^{\lambda x}+b \right )^{2}\right )}{4} \\ &= {\mathrm e}^{\lambda x} a b +{\mathrm e}^{\lambda x} a \lambda +a^{2} {\mathrm e}^{2 \lambda x}+c - \left ({\mathrm e}^{\lambda x} a \lambda \right )-\frac {\left (2 a \,{\mathrm e}^{\lambda x}+b \right )^{2}}{4}\\ &= c -\frac {b^{2}}{4} \end {align*}
Since the Liouville ode invariant does not depend on the independent variable \(x\) then the transformation \begin {align*} y = v \left (x \right ) z \left (x \right )\tag {3} \end {align*}
is used to change the original ode to a constant coefficients ode in \(v\). In (3) the term \(z \left (x \right )\) is given by \begin {align*} z \left (x \right )&={\mathrm e}^{-\left (\int \frac {p \left (x \right )}{2}d x \right )}\\ &= e^{-\int \frac {2 a \,{\mathrm e}^{\lambda x}+b}{2} }\\ &= {\mathrm e}^{-\frac {x b \lambda +2 a \,{\mathrm e}^{\lambda x}}{2 \lambda }}\tag {5} \end {align*}
Hence (3) becomes \begin {align*} y = v \left (x \right ) {\mathrm e}^{-\frac {x b \lambda +2 a \,{\mathrm e}^{\lambda x}}{2 \lambda }}\tag {4} \end {align*}
Applying this change of variable to the original ode results in \begin {align*} -{\mathrm e}^{-\frac {x b \lambda +2 a \,{\mathrm e}^{\lambda x}}{2 \lambda }} \left (b^{2} v \left (x \right )-4 c v \left (x \right )-4 v^{\prime \prime }\left (x \right )\right ) = 0 \end {align*}
Which is now solved for \(v \left (x \right )\) This is second order with constant coefficients homogeneous ODE. In standard form the ODE is \[ A v''(x) + B v'(x) + C v(x) = 0 \] Where in the above \(A=-4, B=0, C=b^{2}-4 c\). Let the solution be \(v \left (x \right )=e^{\lambda x}\). Substituting this into the ODE gives \[ -4 \lambda ^{2} {\mathrm e}^{\lambda x}+\left (b^{2}-4 c \right ) {\mathrm e}^{\lambda x} = 0 \tag {1} \] Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda x}\) gives \[ b^{2}-4 \lambda ^{2}-4 c = 0 \tag {2} \] 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 \(A=-4, B=0, C=b^{2}-4 c\) into the above gives \begin {align*} \lambda _{1,2} &= \frac {0}{(2) \left (-4\right )} \pm \frac {1}{(2) \left (-4\right )} \sqrt {0^2 - (4) \left (-4\right )\left (b^{2}-4 c\right )}\\ &= \pm -\frac {\sqrt {b^{2}-4 c}}{2} \end {align*}
Hence \begin{align*} \lambda _1 &= + -\frac {\sqrt {b^{2}-4 c}}{2} \\ \lambda _2 &= - -\frac {\sqrt {b^{2}-4 c}}{2} \\ \end{align*} Which simplifies to \begin{align*} \lambda _1 &= -\frac {\sqrt {b^{2}-4 c}}{2} \\ \lambda _2 &= \frac {\sqrt {b^{2}-4 c}}{2} \\ \end{align*} Since roots are real and distinct, then the solution is \begin{align*} v \left (x \right ) &= c_{1} e^{\lambda _1 x} + c_{2} e^{\lambda _2 x} \\ v \left (x \right ) &= c_{1} e^{\left (-\frac {\sqrt {b^{2}-4 c}}{2}\right )x} +c_{2} e^{\left (\frac {\sqrt {b^{2}-4 c}}{2}\right )x} \\ \end{align*} Or \[ v \left (x \right ) =c_{1} {\mathrm e}^{-\frac {\sqrt {b^{2}-4 c}\, x}{2}}+c_{2} {\mathrm e}^{\frac {\sqrt {b^{2}-4 c}\, x}{2}} \] Now that \(v \left (x \right )\) is known, then \begin {align*} y&= v \left (x \right ) z \left (x \right )\\ &= \left (c_{1} {\mathrm e}^{-\frac {\sqrt {b^{2}-4 c}\, x}{2}}+c_{2} {\mathrm e}^{\frac {\sqrt {b^{2}-4 c}\, x}{2}}\right ) \left (z \left (x \right )\right )\tag {7} \end {align*}
But from (5) \begin {align*} z \left (x \right )&= {\mathrm e}^{-\frac {x b \lambda +2 a \,{\mathrm e}^{\lambda x}}{2 \lambda }} \end {align*}
Hence (7) becomes \begin {align*} y = \left (c_{1} {\mathrm e}^{-\frac {\sqrt {b^{2}-4 c}\, x}{2}}+c_{2} {\mathrm e}^{\frac {\sqrt {b^{2}-4 c}\, x}{2}}\right ) {\mathrm e}^{-\frac {x b \lambda +2 a \,{\mathrm e}^{\lambda x}}{2 \lambda }} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{1} {\mathrm e}^{-\frac {\sqrt {b^{2}-4 c}\, x}{2}}+c_{2} {\mathrm e}^{\frac {\sqrt {b^{2}-4 c}\, x}{2}}\right ) {\mathrm e}^{-\frac {x b \lambda +2 a \,{\mathrm e}^{\lambda x}}{2 \lambda }} \\ \end{align*}
Verification of solutions
\[ y = \left (c_{1} {\mathrm e}^{-\frac {\sqrt {b^{2}-4 c}\, x}{2}}+c_{2} {\mathrm e}^{\frac {\sqrt {b^{2}-4 c}\, x}{2}}\right ) {\mathrm e}^{-\frac {x b \lambda +2 a \,{\mathrm e}^{\lambda x}}{2 \lambda }} \] Verified OK.
Maple trace Kovacic algorithm successful
`Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Group is reducible or imprimitive <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 54
dsolve(diff(y(x),x$2)+(2*a*exp(lambda*x)+b)*diff(y(x),x)+(a^2*exp(2*lambda*x)+a*(b+lambda)*exp(lambda*x)+c)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{-\frac {b \lambda x +2 \,{\mathrm e}^{x \lambda } a}{2 \lambda }} \left (c_{1} \sinh \left (\frac {\sqrt {b^{2}-4 c}\, x}{2}\right )+c_{2} \cosh \left (\frac {\sqrt {b^{2}-4 c}\, x}{2}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.589 (sec). Leaf size: 82
DSolve[y''[x]+(2*a*Exp[\[Lambda]*x]+b)*y'[x]+(a^2*Exp[2*\[Lambda]*x]+a*(b+\[Lambda])*Exp[\[Lambda]*x]+c)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {\left (c_2 e^{x \sqrt {b^2-4 c}}+c_1 \sqrt {b^2-4 c}\right ) e^{-\frac {a e^{\lambda x}}{\lambda }-\frac {1}{2} x \left (\sqrt {b^2-4 c}+b\right )}}{\sqrt {b^2-4 c}} \]