34.32 problem 32

Internal problem ID [11119]
Internal file name [OUTPUT/10376_Wednesday_January_31_2024_08_14_02_PM_16279652/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: 32.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "exact linear second order ode", "second_order_integrable_as_is"

Maple gives the following as the ode type

[[_2nd_order, _exact, _linear, _homogeneous]]

\[ \boxed {y^{\prime \prime }+\left (a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c \right ) y^{\prime }+\left ({\mathrm e}^{\lambda x} a \lambda +b \mu \,{\mathrm e}^{\mu x}\right ) y=0} \]

34.32.1 Solving as second order integrable as is ode

Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime }+\left (a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c \right ) y^{\prime }+\left ({\mathrm e}^{\lambda x} a \lambda +b \mu \,{\mathrm e}^{\mu x}\right ) y\right )d x &= 0 \\ \left (a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c \right ) y+y^{\prime } = c_{1} \end {align*}

Which is now solved for \(y\).

Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}

Where here \begin {align*} p(x) &=a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c\\ q(x) &=c_{1} \end {align*}

Hence the ode is \begin {align*} \left (a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c \right ) y+y^{\prime } = c_{1} \end {align*}

The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \left (a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c \right )d x} \\ &= {\mathrm e}^{x c +\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }+\frac {b \,{\mathrm e}^{\mu x}}{\mu }} \\ \end{align*} Which simplifies to \[ \mu = {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (c_{1}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left ({\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }} y\right ) &= \left ({\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}\right ) \left (c_{1}\right )\\ \mathrm {d} \left ({\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }} y\right ) &= \left (c_{1} {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}\right )\, \mathrm {d} x \end {align*}

Integrating gives \begin {align*} {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }} y &= \int {c_{1} {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}\,\mathrm {d} x}\\ {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }} y &= \int c_{1} {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}d x + c_{2} \end {align*}

Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}\) results in \begin {align*} y &= {\mathrm e}^{\frac {-a \,{\mathrm e}^{\lambda x} \mu -\lambda \left (c x \mu +b \,{\mathrm e}^{\mu x}\right )}{\mu \lambda }} \left (\int c_{1} {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}d x \right )+c_{2} {\mathrm e}^{\frac {-a \,{\mathrm e}^{\lambda x} \mu -\lambda \left (c x \mu +b \,{\mathrm e}^{\mu x}\right )}{\mu \lambda }} \end {align*}

which simplifies to \begin {align*} y &= {\mathrm e}^{\frac {-a \,{\mathrm e}^{\lambda x} \mu -\lambda \left (c x \mu +b \,{\mathrm e}^{\mu x}\right )}{\mu \lambda }} \left (c_{1} \left (\int {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}d x \right )+c_{2} \right ) \end {align*}

Summary

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

Verification of solutions

\[ y = {\mathrm e}^{\frac {-a \,{\mathrm e}^{\lambda x} \mu -\lambda \left (c x \mu +b \,{\mathrm e}^{\mu x}\right )}{\mu \lambda }} \left (c_{1} \left (\int {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}d x \right )+c_{2} \right ) \] Verified OK.

34.32.2 Solving as type second_order_integrable_as_is (not using ABC version)

Writing the ode as \[ y^{\prime \prime }+\left (a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c \right ) y^{\prime }+\left ({\mathrm e}^{\lambda x} a \lambda +b \mu \,{\mathrm e}^{\mu x}\right ) y = 0 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime }+\left (a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c \right ) y^{\prime }+\left ({\mathrm e}^{\lambda x} a \lambda +b \mu \,{\mathrm e}^{\mu x}\right ) y\right )d x &= 0 \\ y^{\prime }+y c +y \,{\mathrm e}^{\lambda x} a +y \,{\mathrm e}^{\mu x} b = c_{1} \end {align*}

Which is now solved for \(y\).

Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}

Where here \begin {align*} p(x) &=a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c\\ q(x) &=c_{1} \end {align*}

Hence the ode is \begin {align*} \left (a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c \right ) y+y^{\prime } = c_{1} \end {align*}

The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \left (a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c \right )d x} \\ &= {\mathrm e}^{x c +\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }+\frac {b \,{\mathrm e}^{\mu x}}{\mu }} \\ \end{align*} Which simplifies to \[ \mu = {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (c_{1}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left ({\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }} y\right ) &= \left ({\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}\right ) \left (c_{1}\right )\\ \mathrm {d} \left ({\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }} y\right ) &= \left (c_{1} {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}\right )\, \mathrm {d} x \end {align*}

Integrating gives \begin {align*} {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }} y &= \int {c_{1} {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}\,\mathrm {d} x}\\ {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }} y &= \int c_{1} {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}d x + c_{2} \end {align*}

Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}\) results in \begin {align*} y &= {\mathrm e}^{\frac {-a \,{\mathrm e}^{\lambda x} \mu -\lambda \left (c x \mu +b \,{\mathrm e}^{\mu x}\right )}{\mu \lambda }} \left (\int c_{1} {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}d x \right )+c_{2} {\mathrm e}^{\frac {-a \,{\mathrm e}^{\lambda x} \mu -\lambda \left (c x \mu +b \,{\mathrm e}^{\mu x}\right )}{\mu \lambda }} \end {align*}

which simplifies to \begin {align*} y &= {\mathrm e}^{\frac {-a \,{\mathrm e}^{\lambda x} \mu -\lambda \left (c x \mu +b \,{\mathrm e}^{\mu x}\right )}{\mu \lambda }} \left (c_{1} \left (\int {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}d x \right )+c_{2} \right ) \end {align*}

Summary

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

Verification of solutions

\[ y = {\mathrm e}^{\frac {-a \,{\mathrm e}^{\lambda x} \mu -\lambda \left (c x \mu +b \,{\mathrm e}^{\mu x}\right )}{\mu \lambda }} \left (c_{1} \left (\int {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}d x \right )+c_{2} \right ) \] Verified OK.

34.32.3 Solving as exact linear second order ode ode

An ode of the form \begin {align*} p \left (x \right ) y^{\prime \prime }+q \left (x \right ) y^{\prime }+r \left (x \right ) y&=s \left (x \right ) \end {align*}

is exact if \begin {align*} p''(x) - q'(x) + r(x) &= 0 \tag {1} \end {align*}

For the given ode we have \begin {align*} p(x) &= 1\\ q(x) &= a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c\\ r(x) &= {\mathrm e}^{\lambda x} a \lambda +b \mu \,{\mathrm e}^{\mu x}\\ s(x) &= 0 \end {align*}

Hence \begin {align*} p''(x) &= 0\\ q'(x) &= {\mathrm e}^{\lambda x} a \lambda +b \mu \,{\mathrm e}^{\mu x} \end {align*}

Therefore (1) becomes \begin {align*} 0- \left ({\mathrm e}^{\lambda x} a \lambda +b \mu \,{\mathrm e}^{\mu x}\right ) + \left ({\mathrm e}^{\lambda x} a \lambda +b \mu \,{\mathrm e}^{\mu x}\right )&=0 \end {align*}

Hence the ode is exact. Since we now know the ode is exact, it can be written as \begin {align*} \left (p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y\right )' &= s(x) \end {align*}

Integrating gives \begin {align*} p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y&=\int {s \left (x \right )\, dx} \end {align*}

Substituting the above values for \(p,q,r,s\) gives \begin {align*} \left (a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c \right ) y+y^{\prime }&=c_{1} \end {align*}

We now have a first order ode to solve which is \begin {align*} \left (a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c \right ) y+y^{\prime } = c_{1} \end {align*}

Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}

Where here \begin {align*} p(x) &=a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c\\ q(x) &=c_{1} \end {align*}

Hence the ode is \begin {align*} \left (a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c \right ) y+y^{\prime } = c_{1} \end {align*}

The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \left (a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c \right )d x} \\ &= {\mathrm e}^{x c +\frac {a \,{\mathrm e}^{\lambda x}}{\lambda }+\frac {b \,{\mathrm e}^{\mu x}}{\mu }} \\ \end{align*} Which simplifies to \[ \mu = {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (c_{1}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left ({\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }} y\right ) &= \left ({\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}\right ) \left (c_{1}\right )\\ \mathrm {d} \left ({\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }} y\right ) &= \left (c_{1} {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}\right )\, \mathrm {d} x \end {align*}

Integrating gives \begin {align*} {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }} y &= \int {c_{1} {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}\,\mathrm {d} x}\\ {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }} y &= \int c_{1} {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}d x + c_{2} \end {align*}

Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}\) results in \begin {align*} y &= {\mathrm e}^{\frac {-a \,{\mathrm e}^{\lambda x} \mu -\lambda \left (c x \mu +b \,{\mathrm e}^{\mu x}\right )}{\mu \lambda }} \left (\int c_{1} {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}d x \right )+c_{2} {\mathrm e}^{\frac {-a \,{\mathrm e}^{\lambda x} \mu -\lambda \left (c x \mu +b \,{\mathrm e}^{\mu x}\right )}{\mu \lambda }} \end {align*}

which simplifies to \begin {align*} y &= {\mathrm e}^{\frac {-a \,{\mathrm e}^{\lambda x} \mu -\lambda \left (c x \mu +b \,{\mathrm e}^{\mu x}\right )}{\mu \lambda }} \left (c_{1} \left (\int {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}d x \right )+c_{2} \right ) \end {align*}

Summary

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

Verification of solutions

\[ y = {\mathrm e}^{\frac {-a \,{\mathrm e}^{\lambda x} \mu -\lambda \left (c x \mu +b \,{\mathrm e}^{\mu x}\right )}{\mu \lambda }} \left (c_{1} \left (\int {\mathrm e}^{\frac {x c \lambda \mu +a \,{\mathrm e}^{\lambda x} \mu +b \,{\mathrm e}^{\mu x} \lambda }{\lambda \mu }}d x \right )+c_{2} \right ) \] Verified OK.

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a symmetry of the form [xi=0, eta=F(x)] 
   One independent solution has integrals. Trying a hypergeometric solution free of integrals... 
   -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
No hypergeometric solution was found. 
<- linear_1 successful`
 

✓ Solution by Maple

Time used: 0.015 (sec). Leaf size: 70

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

\[ y \left (x \right ) = \left (c_{1} \left (\int {\mathrm e}^{\frac {c x \mu \lambda +{\mathrm e}^{x \lambda } a \mu +b \,{\mathrm e}^{x \mu } \lambda }{\mu \lambda }}d x \right )+c_{2} \right ) {\mathrm e}^{\frac {-{\mathrm e}^{x \lambda } a \mu -\lambda \left (c x \mu +b \,{\mathrm e}^{x \mu }\right )}{\mu \lambda }} \]

✓ Solution by Mathematica

Time used: 0.146 (sec). Leaf size: 77

DSolve[y''[x]+(a*Exp[\[Lambda]*x]+b*Exp[\[Mu]*x]+c)*y'[x]+(a*\[Lambda]*Exp[\[Lambda]*x]+b*\[Mu]*Exp[\[Mu]*x])*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to e^{-\frac {a e^{\lambda x}}{\lambda }-\frac {b e^{\mu x}}{\mu }-c x} \left (\int _1^xe^{\frac {e^{\lambda K[1]} a}{\lambda }+c K[1]+\frac {b e^{\mu K[1]}}{\mu }} c_1dK[1]+c_2\right ) \]