4.3.2.15 Bessel form A type ode $a{y}^{\prime \prime }+b{y}^{\prime }+(c{e}^{rx}-m)y=f\left(x\right)$

ode internal name "second_order_bessel_ode_form_A"

These are ode of the above form which can be converted to Bessel using transformation $x=\ln \left(t\right)$.

4.3.2.15.1 Example $a{y}^{\prime \prime }+b{y}^{\prime }+(c{e}^{rx}-m)y=0$ An ode of the form

$$\begin{array}{}\text{(1)}& a{y}^{\prime \prime }+b{y}^{\prime }+(c{e}^{rx}+m)y=0\end{array}$$

can be transformed to Bessel ode using the transformation

$$\begin{array}{rl}x& =\ln \left(t\right)\\ e^x& =t\end{array}$$

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

$$\begin{array}{rl}\frac{dy}{dx}& =\frac{dy}{dt}\frac{dt}{dx}\\ & =\frac{dy}{dt}{e}^x\\ \text{(2)}& & =t\frac{dy}{dt}\end{array}$$

And

$$\begin{array}{rl}\frac{d^{2}y}{d{x}^2}& =\frac{d}{dx}\left(\frac{dy}{dx}\right)\\ & =\frac{d}{dx}\left(t\frac{dy}{dt}\right)\\ & =\frac{d}{dt}\frac{dt}{dx}\left(t\frac{dy}{dt}\right)\\ & =\frac{dt}{dx}\frac{d}{dt}\left(t\frac{dy}{dt}\right)\\ & =t\frac{d}{dt}\left(t\frac{dy}{dt}\right)\\ \text{(3)}& & =t(\frac{dy}{dt}+t\frac{d^{2}y}{d{t}^2})\end{array}$$

Substituting (2,3) into (1) gives

$$\begin{array}{rl}at(\frac{dy}{dt}+t\frac{d^{2}y}{d{t}^2})+bt\frac{dy}{dt}+(c{e}^{rx}+m)y& =0\\ (at{y}^{\prime }+a{t}^2{y}^{\prime \prime })+bt{y}^{\prime }+(c{t}^r+m)y& =0\\ a{t}^2{y}^{\prime \prime }+(b+a)t{y}^{\prime }+(c{t}^r+m)y& =0\\ \text{(4)}& t^2{y}^{\prime \prime }+\frac{b+a}{a}t{y}^{\prime }+(\frac{c}{a}{t}^r+\frac{m}{a})y& =0\end{array}$$

Which is Bessel ODE. Comparing the above to the general known Bowman form of Bessel ode which is

$$\begin{array}{}\text{(C)}& t^2{y}^{\prime \prime }+(1-2\alpha )t{y}^{\prime }+({\beta }^2{\gamma }^2{t}^{2\gamma }-(n^2{\gamma }^2-{\alpha }^2))y=0\end{array}$$

And now comparing (4) and (C) shows that

$$\begin{array}{}\text{(5)}& (1-2\alpha )& =\frac{b+a}{a}\text{(6)}& {\beta }^2{\gamma }^2& =\frac{c}{a}\text{(7)}& 2\gamma & =r\text{(8)}& (n^2{\gamma }^2-{\alpha }^2)& =-\frac{m}{a}\end{array}$$

(5) gives $\alpha =\frac{1}{2}-\frac{b+a}{2a}$. (7) gives $\gamma =\frac{r}{2}$. (8) now becomes $(n^2{\left(\frac{r}{2}\right)}^2-{(\frac{1}{2}-\frac{b+a}{2a})}^2)=-\frac{m}{a}$ or $n^2=\frac{-\frac{m}{a}+{(\frac{1}{2}-\frac{b+a}{2a})}^2}{{\left(\frac{r}{2}\right)}^2}$. Hence $n=\frac{2}{r}\sqrt{-\frac{m}{a}+{(\frac{1}{2}-\frac{b+a}{2a})}^2}$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

$$\begin{array}{rl}\alpha & =\frac{1}{2}-\frac{b+a}{2a}\\ n& =\frac{2}{r}\sqrt{-\frac{m}{a}+{(\frac{1}{2}-\frac{b+a}{2a})}^2}\\ \beta & =\sqrt{\frac{c}{a}}\frac{2}{r}\\ \gamma & =\frac{r}{2}\end{array}$$

But the solution to (C) which is general form of Bessel ode is known and given by

$$y\left(t\right)=t^{\alpha }(c_1{J}_n\left(\beta t^{\gamma }\right)+c_2{Y}_n\left(\beta t^{\gamma }\right))$$

Substituting the above values found into this solution gives

$$y\left(t\right)=t^{\frac{1}{2}-\frac{b+a}{2a}}(c_1{J}_{\frac{2}{r}\sqrt{-\frac{m}{a}+{(\frac{1}{2}-\frac{b+a}{2a})}^2}}\left(\sqrt{\frac{c}{a}}\frac{2}{r}{t}^{\frac{r}{2}}\right)+c_2{Y}_{\frac{2}{r}\sqrt{-\frac{m}{a}+{(\frac{1}{2}-\frac{b+a}{2a})}^2}}\left(\sqrt{\frac{c}{a}}\frac{2}{r}{t}^{\frac{r}{2}}\right))$$

Since $e^x=t$ then the above becomes

$$\begin{array}{rl}y\left(x\right)& =e^{x(\frac{1}{2}-\frac{b+a}{2a})}(c_1{J}_{\frac{2}{r}\sqrt{-\frac{m}{a}+{(\frac{1}{2}-\frac{b+a}{2a})}^2}}\left(\sqrt{\frac{c}{a}}\frac{2}{r}{e}^{x\frac{r}{2}}\right)+c_2{Y}_{\frac{2}{r}\sqrt{-\frac{m}{a}+{(\frac{1}{2}-\frac{b+a}{2a})}^2}}\left(\sqrt{\frac{c}{a}}\frac{2}{r}{e}^{x\frac{r}{2}}\right))\\ & =e^{x\left(\frac{-b}{2a}\right)}(c_1{J}_{\frac{2}{r}\sqrt{-\frac{m}{a}+{\left(\frac{-b}{2a}\right)}^2}}\left(\sqrt{\frac{c}{a}}\frac{2}{r}{e}^{x\frac{r}{2}}\right)+c_2{Y}_{\frac{2}{r}\sqrt{-\frac{m}{a}+{\left(\frac{-b}{2a}\right)}^2}}\left(\sqrt{\frac{c}{a}}\frac{2}{r}{e}^{x\frac{r}{2}}\right))\\ & =e^{x\left(\frac{-b}{2a}\right)}(c_1{J}_{\frac{2}{r}\sqrt{-\frac{m}{a}+\frac{b^2}{4a^2}}}\left(\sqrt{\frac{c}{a}}\frac{2}{r}{e}^{x\frac{r}{2}}\right)+c_2{Y}_{\frac{2}{r}\sqrt{-\frac{m}{a}+\frac{b^2}{4a^2}}}\left(\sqrt{\frac{c}{a}}\frac{2}{r}{e}^{x\frac{r}{2}}\right))\\ & =e^{x\left(\frac{-b}{2a}\right)}(c_1{J}_{\frac{2}{r}\sqrt{-\frac{4ma+b^2}{4a^2}}}\left(\sqrt{\frac{c}{a}}\frac{2}{r}{e}^{x\frac{r}{2}}\right)+c_2{Y}_{\frac{2}{r}\sqrt{-\frac{4ma+b^2}{4a^2}}}\left(\sqrt{\frac{c}{a}}\frac{2}{r}{e}^{x\frac{r}{2}}\right))\\ \text{(9)}& & =e^{x\left(\frac{-b}{2a}\right)}(c_1{J}_{\frac{1}{ra}\sqrt{-4ma+b^2}}\left(\sqrt{\frac{c}{a}}\frac{2}{r}{e}^{x\frac{r}{2}}\right)+c_2{Y}_{\frac{1}{ra}\sqrt{-4ma+b^2}}\left(\sqrt{\frac{c}{a}}\frac{2}{r}{e}^{x\frac{r}{2}}\right))\end{array}$$

Equation (9) above is the solution to $a{y}^{\prime \prime }+b{y}^{\prime }+(c{e}^{rx}+m)y=0$. Therefore we just need now to compare this form to the ode given and use (9) to obtain the final solution. Let us now apply this to an example for illustration. Given the ode

$$y^{\prime \prime }+(e^{2x}-4)y=0$$

Comparing the above to $a{y}^{\prime \prime }+b{y}^{\prime }+(c{e}^{rx}+m)y=0$ shows that $a=1,b=0,c=1,r=2,m=-4$. Hence the solution (9) becomes

$$\begin{array}{rl}y\left(x\right)& =e^{x\left(\frac{-b}{2a}\right)}(c_1{J}_{\frac{1}{ra}\sqrt{-4ma+b^2}}\left(\sqrt{\frac{c}{a}}\frac{2}{r}{e}^{x\frac{r}{2}}\right)+c_2{Y}_{\frac{1}{ra}\sqrt{-4ma+b^2}}\left(\sqrt{\frac{c}{a}}\frac{2}{r}{e}^{x\frac{r}{2}}\right))\\ & =c_1{J}_{\frac{1}{2}\sqrt{16}}\left(e^x\right)+c_2{Y}_{\frac{1}{2}\sqrt{16}}\left(e^x\right)\\ & =c_1{J}_2\left(e^x\right)+c_2{Y}_2\left(e^x\right)\\ & =c_1\mathrm{BesselJ}(2,e^x)+c_2\mathrm{BesselY}(2,e^x)\end{array}$$

Another example for illustration. Given the ode

$$y^{\prime \prime }+y^{\prime }+(e^x-4)y=0$$

Comparing the above to $a{y}^{\prime \prime }+b{y}^{\prime }+(c{e}^{rx}+m)y=0$ shows that $a=1,b=1,c=1,r=1,m=-4$. Hence the solution (9) becomes

$$\begin{array}{rl}y\left(x\right)& =e^{x\left(\frac{-1}{2}\right)}(c_1{J}_{\sqrt{16}}\left(2e^{x\frac{1}{2}}\right)+c_2{Y}_{\sqrt{16+1}}\left(2e^{x\frac{1}{2}}\right))\\ & =e^{\frac{-x}{2}}(c_1{J}_{\sqrt{17}}\left(2e^{\frac{x}{2}}\right)+c_2{Y}_{\sqrt{17}}\left(2e^{\frac{x}{2}}\right))\end{array}$$

Another example for illustration. Given the ode

$$y^{\prime \prime }+(e^{2x}-n^2)y=0$$

Comparing the above to $a{y}^{\prime \prime }+b{y}^{\prime }+(c{e}^{rx}+m)y=0$ shows that $a=1,b=0,c=1,r=2,m=-n^2$. Hence the solution (9) becomes

$$\begin{array}{rl}y\left(x\right)& =e^{x\left(\frac{-b}{2a}\right)}(c_1{J}_{\frac{1}{ra}\sqrt{-4ma+b^2}}\left(\sqrt{\frac{c}{a}}\frac{2}{r}{e}^{x\frac{r}{2}}\right)+c_2{Y}_{\frac{1}{ra}\sqrt{-4ma+b^2}}\left(\sqrt{\frac{c}{a}}\frac{2}{r}{e}^{x\frac{r}{2}}\right))\\ & =c_1{J}_{\frac{1}{2}\sqrt{-4(-n^2)}}\left(e^x\right)+c_2{Y}_{\frac{1}{2}\sqrt{-4(-n^2)}}\left(e^x\right)\\ & =c_1{J}_n\left(e^x\right)+c_2{Y}_n\left(e^x\right)\\ & =c_1\mathrm{BesselJ}(n,e^x)+c_2\mathrm{BesselY}(n,e^x)\end{array}$$