Internal problem ID [6346]
Internal file name [OUTPUT/5594_Sunday_June_05_2022_03_44_24_PM_57895568/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 2. Second-Order Linear Equations. Section 2.4. THE USE OF A KNOWN
SOLUTION TO FIND ANOTHER. Page 74
Problem number: 7.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_change_of_variable_on_y_method_2"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime }-x f \left (x \right ) y^{\prime }+f \left (x \right ) y=0} \]
In normal form the ode \begin {align*} y^{\prime \prime }-x f \left (x \right ) y^{\prime }+f \left (x \right ) 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 )&=-f \left (x \right ) x\\ q \left (x \right )&=f \left (x \right ) \end {align*}
Applying change of variables on the depndent variable \(y = v \left (x \right ) x^{n}\) to (2) gives the following ode where the dependent variables is \(v \left (x \right )\) and not \(y\). \begin {align*} v^{\prime \prime }\left (x \right )+\left (\frac {2 n}{x}+p \right ) v^{\prime }\left (x \right )+\left (\frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q \right ) v \left (x \right )&=0 \tag {3} \end {align*}
Let the coefficient of \(v \left (x \right )\) above be zero. Hence \begin {align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q&=0 \tag {4} \end {align*}
Substituting the earlier values found for \(p \left (x \right )\) and \(q \left (x \right )\) into (4) gives \begin {align*} \frac {n \left (n -1\right )}{x^{2}}-n f \left (x \right )+f \left (x \right )&=0 \tag {5} \end {align*}
Solving (5) for \(n\) gives \begin {align*} n&=1 \tag {6} \end {align*}
Substituting this value in (3) gives \begin {align*} v^{\prime \prime }\left (x \right )+\left (\frac {2}{x}-f \left (x \right ) x \right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\left (\frac {2}{x}-f \left (x \right ) x \right ) v^{\prime }\left (x \right )&=0 \tag {7} \\ \end {align*}
Using the substitution \begin {align*} u \left (x \right ) = v^{\prime }\left (x \right ) \end {align*}
Then (7) becomes \begin {align*} u^{\prime }\left (x \right )+\left (\frac {2}{x}-f \left (x \right ) x \right ) u \left (x \right ) = 0 \tag {8} \\ \end {align*}
The above is now solved for \(u \left (x \right )\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {u \left (f \left (x \right ) x^{2}-2\right )}{x} \end {align*}
Where \(f(x)=\frac {f \left (x \right ) x^{2}-2}{x}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= \frac {f \left (x \right ) x^{2}-2}{x} \,d x\\ \int { \frac {1}{u} \,du} &= \int {\frac {f \left (x \right ) x^{2}-2}{x} \,d x}\\ \ln \left (u \right )&=\int \frac {f \left (x \right ) x^{2}-2}{x}d x +c_{1}\\ u&={\mathrm e}^{\int \frac {f \left (x \right ) x^{2}-2}{x}d x +c_{1}}\\ &=c_{1} {\mathrm e}^{\int \frac {f \left (x \right ) x^{2}-2}{x}d x} \end {align*}
Now that \(u \left (x \right )\) is known, then \begin {align*} v^{\prime }\left (x \right )&= u \left (x \right )\\ v \left (x \right )&= \int u \left (x \right )d x +c_{2}\\ &= \int c_{1} {\mathrm e}^{\int \frac {f \left (x \right ) x^{2}-2}{x}d x}d x +c_{2} \end {align*}
Hence \begin {align*} y&= v \left (x \right ) x^{n}\\ &= \left (\int c_{1} {\mathrm e}^{\int \frac {f \left (x \right ) x^{2}-2}{x}d x}d x +c_{2} \right ) x\\ &= \left (c_{1} \left (\int {\mathrm e}^{\int \frac {f \left (x \right ) x^{2}-2}{x}d x}d x \right )+c_{2} \right ) x\\ \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (\int c_{1} {\mathrm e}^{\int \frac {f \left (x \right ) x^{2}-2}{x}d x}d x +c_{2} \right ) x \\ \end{align*}
Verification of solutions
\[ y = \left (\int c_{1} {\mathrm e}^{\int \frac {f \left (x \right ) x^{2}-2}{x}d x}d x +c_{2} \right ) x \] Verified OK.
Maple trace
`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 -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) -> Trying changes of variables to rationalize or make the ODE simpler <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) <- linear symmetries successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 27
dsolve(diff(y(x),x$2)-x*f(x)*diff(y(x),x)+f(x)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = x \left (\left (\int {\mathrm e}^{\int \frac {-2+f \left (x \right ) x^{2}}{x}d x}d x \right ) c_{1} +c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 0.287 (sec). Leaf size: 44
DSolve[y''[x]-x*f[x]*y'[x]+f[x]*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to x \left (c_2 \int _1^x\frac {\exp \left (-\int _1^{K[2]}-f(K[1]) K[1]dK[1]\right )}{K[2]^2}dK[2]+c_1\right ) \]