2.1102   ODE No. 1102

\[ a x^2 y(x)+x y''(x)+2 y'(x)=0 \]

✓ Mathematica : cpu = 0.0049484 (sec), leaf count = 42

DSolve[a*x^2*y[x] + 2*Derivative[1][y][x] + x*Derivative[2][y][x] == 0,y[x],x]
 

\[\left \{\left \{y(x)\to \frac {c_1 \text {Ai}\left (-\frac {a x}{(-a)^{2/3}}\right )}{x}+\frac {c_2 \text {Bi}\left (-\frac {a x}{(-a)^{2/3}}\right )}{x}\right \}\right \}\]

✓ Maple : cpu = 0.042 (sec), leaf count = 33

dsolve(x*diff(diff(y(x),x),x)+2*diff(y(x),x)+y(x)*a*x^2=0,y(x))
 

\[y \left (x \right ) = \frac {c_{2} \operatorname {BesselY}\left (\frac {1}{3}, \frac {2 \sqrt {a}\, x^{{3}/{2}}}{3}\right )+c_{1} \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 \sqrt {a}\, x^{{3}/{2}}}{3}\right )}{\sqrt {x}}\]

Hand solution

\begin{equation} xy^{\prime \prime }+2y^{\prime }+ax^{2}y=0\tag {1}\end{equation}

Since there is a term \(2y\), we can use \(y=\frac {u\left ( x\right ) }{x}\), hence

\begin{align*} y^{\prime } & =\frac {u^{\prime }}{x}-\frac {u}{x^{2}}\\ y^{\prime \prime } & =\frac {u^{\prime \prime }}{x}-\frac {u^{\prime }}{x^{2}}-\frac {u^{\prime }}{x^{2}}+2\frac {u}{x^{3}}\end{align*}

And (1) becomes

\begin{align} x\left ( \frac {u^{\prime \prime }}{x}-\frac {u^{\prime }}{x^{2}}-\frac {u^{\prime }}{x^{2}}+2\frac {u}{x^{3}}\right ) +2\left ( \frac {u^{\prime }}{x}-\frac {u}{x^{2}}\right ) +ax^{2}\left ( \frac {u}{x}\right ) & =0\nonumber \\ u^{\prime \prime }-2\frac {u^{\prime }}{x}+2\frac {u}{x^{2}}+\frac {2u^{\prime }}{x}-\frac {2u}{x^{2}}+axu & =0\nonumber \\ u^{\prime \prime }+axu & =0\tag {2}\end{align}

This is Emdon-Fowler. (form is \(u^{\prime \prime }+x^{n}u=0\)) with \(n=1\).  Assume that

\[ u=\sum _{n=0}^{\infty }c_{n}x^{n}\]

Hence

\begin{align*} u^{\prime } & =\sum _{n=0}nc_{n}x^{n-1}=\sum _{n=1}nc_{n}x^{n-1}=\sum _{n=0}\left ( n+1\right ) c_{n+1}x^{n}\\ u^{\prime \prime } & =\sum _{n=0}n\left ( n+1\right ) c_{n+1}x^{n-1}=\sum _{n=1}n\left ( n+1\right ) c_{n+1}x^{n-1}=\sum _{n=0}\left ( n+1\right ) \left ( n+2\right ) c_{n+2}x^{n}\end{align*}

Substituting back in (2) gives

\begin{align*} \sum _{n=0}\left ( n+1\right ) \left ( n+2\right ) c_{n+2}x^{n}+\sum _{n=0}ac_{n}x^{n+1} & =0\\ \sum _{n=0}\left ( n+1\right ) \left ( n+2\right ) c_{n+2}x^{n}+\sum _{n=1}ac_{n-1}x^{n} & =0 \end{align*}

For \(n=0\)

\[ \left ( 1\right ) \left ( 2\right ) c_{2}=0 \]

Hence \(c_{2}=0\).  For \(n\geq 1\)

\begin{align} \left ( n+1\right ) \left ( n+2\right ) c_{n+2}+ac_{n-1} & =0\nonumber \\ c_{n+2} & =\frac {-ac_{n-1}}{\left ( n+1\right ) \left ( n+2\right ) }\tag {3}\end{align}

For \(n=1\,\), from (3)

\[ c_{3}=\frac {-ac_{0}}{\left ( 2\right ) \left ( 3\right ) }\]

For \(n=2\), from (3)

\[ c_{4}=\frac {-ac_{1}}{\left ( 3\right ) \left ( 4\right ) }\]

For \(n=3\), from (3)

\[ c_{5}=\frac {-ac_{2}}{\left ( 4\right ) \left ( 5\right ) }=0 \]

For \(n=4\), from (3)

\[ c_{6}=\frac {-ac_{3}}{\left ( 5\right ) \left ( 6\right ) }=\frac {-a}{\left ( 5\right ) \left ( 6\right ) }\left ( \frac {-ac_{0}}{\left ( 2\right ) \left ( 3\right ) }\right ) =\frac {a^{2}c_{0}}{\left ( 2\right ) \left ( 3\right ) \left ( 5\right ) \left ( 6\right ) }\]

For \(n=5\), from (3)

\[ c_{7}=\frac {-ac_{4}}{\left ( 6\right ) \left ( 7\right ) }=\frac {-a}{\left ( 6\right ) \left ( 7\right ) }\left ( \frac {-ac_{1}}{\left ( 3\right ) \left ( 4\right ) }\right ) =\frac {a^{2}c_{1}}{\left ( 3\right ) \left ( 4\right ) \left ( 6\right ) \left ( 7\right ) }\]

For \(n=6\), from (3)

\[ c_{8}=\frac {-ac_{5}}{\left ( 7\right ) \left ( 8\right ) }=0 \]

For \(n=7\), from (3)

\[ c_{9}=\frac {-ac_{6}}{\left ( 8\right ) \left ( 9\right ) }=\frac {-a}{\left ( 8\right ) \left ( 9\right ) }\left ( \frac {a^{2}c_{0}}{\left ( 2\right ) \left ( 3\right ) \left ( 5\right ) \left ( 6\right ) }\right ) =\frac {-a^{3}c_{0}}{\left ( 2\right ) \left ( 3\right ) \left ( 5\right ) \left ( 6\right ) \left ( 8\right ) \left ( 9\right ) }\]

For \(n=8\), from (3)

\[ c_{10}=\frac {-ac_{7}}{\left ( 9\right ) \left ( 10\right ) }=\frac {-a}{\left ( 9\right ) \left ( 10\right ) }\left ( \frac {a^{2}c_{1}}{\left ( 3\right ) \left ( 4\right ) \left ( 6\right ) \left ( 7\right ) }\right ) =\frac {-a^{3}c_{1}}{\left ( 3\right ) \left ( 4\right ) \left ( 6\right ) \left ( 7\right ) \left ( 9\right ) \left ( 10\right ) }\]

And so on. Hence,

\begin{align*} u & =\sum _{n=0}^{\infty }c_{n}x^{n}=c_{0}+c_{1}x^{1}+c_{2}x^{2}+c_{3}x^{3}+\cdots \\ & =c_{0}+c_{1}x+0+c_{3}x^{3}+c_{4}x^{4}+0+c_{6}x^{6}+c_{7}x^{7}+0+c_{9}x^{9}+\cdots \\ & =c_{0}+c_{1}x-\frac {ac_{0}}{\left ( 2\right ) \left ( 3\right ) }x^{3}-\frac {ac_{1}}{\left ( 3\right ) \left ( 4\right ) }x^{4}+\frac {a^{2}c_{0}}{\left ( 2\right ) \left ( 3\right ) \left ( 5\right ) \left ( 6\right ) }x^{6}+\frac {a^{2}c_{1}}{\left ( 3\right ) \left ( 4\right ) \left ( 6\right ) \left ( 7\right ) }x^{7}-\frac {a^{3}c_{0}}{\left ( 2\right ) \left ( 3\right ) \left ( 5\right ) \left ( 6\right ) \left ( 8\right ) \left ( 9\right ) }x^{9}-\frac {a^{3}c_{1}}{\left ( 3\right ) \left ( 4\right ) \left ( 6\right ) \left ( 7\right ) \left ( 9\right ) \left ( 10\right ) }x^{10}+\cdots \\ & =c_{0}\left ( 1-\frac {a}{6}x^{3}+\frac {a^{2}}{180}x^{6}-\frac {a^{3}}{12\,960}x^{9}+\cdots \right ) +xc_{1}\left ( 1-\frac {a}{12}x^{3}+\frac {a^{2}}{504}x^{6}-\frac {a^{3}}{45\,360}x^{9}+\cdots \right ) \\ & =c_{0}\left ( 1-\frac {1}{6}\left ( a^{\frac {1}{3}}x\right ) ^{3}+\frac {1}{180}\left ( a^{\frac {1}{3}}x\right ) ^{6}-\frac {1}{12\,960}\left ( a^{\frac {1}{3}}x\right ) ^{9}+\cdots \right ) +xc_{1}\left ( 1-\frac {1}{12}\left ( a^{\frac {1}{3}}x\right ) ^{3}+\frac {1}{504}\left ( a^{\frac {1}{3}}x\right ) ^{6}-\frac {1}{45\,360}\left ( a^{\frac {1}{3}}x\right ) ^{9}+\cdots \right ) \end{align*}

Comparing the above to the series expansion of Airy functions, we see that

\[ u=c_{0}\operatorname {AiryAI}\left ( -a^{\frac {1}{3}}x\right ) +c_{1}\operatorname {AiryBI}\left ( -a^{\frac {1}{3}}x\right ) \]

And since \(y=\frac {u\left ( x\right ) }{x}\) then

\[ y=\frac {1}{x}\left ( c_{0}\operatorname {AiryAI}\left ( -a^{\frac {1}{3}}x\right ) +c_{1}\operatorname {AiryBI}\left ( -a^{\frac {1}{3}}x\right ) \right ) \]

Verification

restart; 
#for series solution of u''+axu=0, use this: 
Order:=10; 
sol:=dsolve(ode,u(x),series); 
sol:=convert(sol,polynom); 
sol:=subs({u(0)=c0,D(u)(0)=c1},rhs(sol)); 
collect(sol,{c0,c1}); 
(1-(1/6)*a*x^3+(1/180)*a^2*x^6-(1/12960)*a^3*x^9)*c0+(x-(1/12)*a*x^4+(1/504)*a^2*x^7)*c1 
#to verify final solution use this: 
ode:=x*diff(y(x),x$2)+2*diff(y(x),x)+a*x^2*y(x)=0; 
y0:=(1/x)*(_C0*AiryAi(-a^(1/3)*x)+_C1*AiryBi(-a^(1/3)*x)); 
odetest(y(x)=y0,ode); 
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