Internal problem ID [9588]
Internal file name [OUTPUT/8529_Monday_June_06_2022_03_39_07_AM_25637246/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1260.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_y]]
\[ \boxed {x \left (x -1\right ) y^{\prime \prime }+\left (\left (\operatorname {a1} +\operatorname {b1} +1\right ) x -\operatorname {d1} \right ) y^{\prime }=-\operatorname {a1} \operatorname {b1} \operatorname {d1}} \]
This is second order ode with missing dependent variable \(y\). Let \begin {align*} p(x) &= y^{\prime } \end {align*}
Then \begin {align*} p'(x) &= y^{\prime \prime } \end {align*}
Hence the ode becomes \begin {align*} \left (x^{2}-x \right ) p^{\prime }\left (x \right )+\left (\left (\operatorname {a1} +\operatorname {b1} +1\right ) x -\operatorname {d1} \right ) p \left (x \right )+\operatorname {a1} \operatorname {b1} \operatorname {d1} = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode.
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} p^{\prime }\left (x \right ) + p(x)p \left (x \right ) &= q(x) \end {align*}
Where here \begin {align*} p(x) &=\frac {x \operatorname {a1} +x \operatorname {b1} -\operatorname {d1} +x}{x \left (x -1\right )}\\ q(x) &=-\frac {\operatorname {a1} \operatorname {b1} \operatorname {d1}}{x \left (x -1\right )} \end {align*}
Hence the ode is \begin {align*} p^{\prime }\left (x \right )+\frac {\left (x \operatorname {a1} +x \operatorname {b1} -\operatorname {d1} +x \right ) p \left (x \right )}{x \left (x -1\right )} = -\frac {\operatorname {a1} \operatorname {b1} \operatorname {d1}}{x \left (x -1\right )} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \frac {x \operatorname {a1} +x \operatorname {b1} -\operatorname {d1} +x}{x \left (x -1\right )}d x} \\ &= {\mathrm e}^{\operatorname {d1} \ln \left (x \right )+\left (\operatorname {a1} +\operatorname {b1} +1-\operatorname {d1} \right ) \ln \left (x -1\right )} \\ \end{align*} Which simplifies to \[ \mu = x^{\operatorname {d1}} \left (x -1\right )^{\operatorname {a1} +\operatorname {b1} +1-\operatorname {d1}} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu p\right ) &= \left (\mu \right ) \left (-\frac {\operatorname {a1} \operatorname {b1} \operatorname {d1}}{x \left (x -1\right )}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (x^{\operatorname {d1}} \left (x -1\right )^{\operatorname {a1} +\operatorname {b1} +1-\operatorname {d1}} p\right ) &= \left (x^{\operatorname {d1}} \left (x -1\right )^{\operatorname {a1} +\operatorname {b1} +1-\operatorname {d1}}\right ) \left (-\frac {\operatorname {a1} \operatorname {b1} \operatorname {d1}}{x \left (x -1\right )}\right )\\ \mathrm {d} \left (x^{\operatorname {d1}} \left (x -1\right )^{\operatorname {a1} +\operatorname {b1} +1-\operatorname {d1}} p\right ) &= \left (-\operatorname {a1} \operatorname {b1} \operatorname {d1} \,x^{-1+\operatorname {d1}} \left (x -1\right )^{\operatorname {a1} +\operatorname {b1} -\operatorname {d1}}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} x^{\operatorname {d1}} \left (x -1\right )^{\operatorname {a1} +\operatorname {b1} +1-\operatorname {d1}} p &= \int {-\operatorname {a1} \operatorname {b1} \operatorname {d1} \,x^{-1+\operatorname {d1}} \left (x -1\right )^{\operatorname {a1} +\operatorname {b1} -\operatorname {d1}}\,\mathrm {d} x}\\ x^{\operatorname {d1}} \left (x -1\right )^{\operatorname {a1} +\operatorname {b1} +1-\operatorname {d1}} p &= -\operatorname {a1} \operatorname {b1} \left (-1\right )^{-\operatorname {a1} -\operatorname {b1} +\operatorname {d1}} x^{\operatorname {d1}} \operatorname {hypergeom}\left (\left [\operatorname {d1} , -\operatorname {a1} -\operatorname {b1} +\operatorname {d1} \right ], \left [\operatorname {d1} +1\right ], x\right ) + c_{1} \end {align*}
Dividing both sides by the integrating factor \(\mu =x^{\operatorname {d1}} \left (x -1\right )^{\operatorname {a1} +\operatorname {b1} +1-\operatorname {d1}}\) results in \begin {align*} p \left (x \right ) &= -x^{-\operatorname {d1}} \left (x -1\right )^{-\operatorname {a1} -\operatorname {b1} +\operatorname {d1} -1} \operatorname {a1} \operatorname {b1} \left (-1\right )^{-\operatorname {a1} -\operatorname {b1} +\operatorname {d1}} x^{\operatorname {d1}} \operatorname {hypergeom}\left (\left [\operatorname {d1} , -\operatorname {a1} -\operatorname {b1} +\operatorname {d1} \right ], \left [\operatorname {d1} +1\right ], x\right )+c_{1} x^{-\operatorname {d1}} \left (x -1\right )^{-\operatorname {a1} -\operatorname {b1} +\operatorname {d1} -1} \end {align*}
which simplifies to \begin {align*} p \left (x \right ) &= -\left (x -1\right )^{-\operatorname {a1} -\operatorname {b1} +\operatorname {d1} -1} \left (\left (-1\right )^{-\operatorname {a1} -\operatorname {b1} +\operatorname {d1}} \operatorname {hypergeom}\left (\left [\operatorname {d1} , -\operatorname {a1} -\operatorname {b1} +\operatorname {d1} \right ], \left [\operatorname {d1} +1\right ], x\right ) \operatorname {a1} \operatorname {b1} -x^{-\operatorname {d1}} c_{1} \right ) \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = -\left (x -1\right )^{-\operatorname {a1} -\operatorname {b1} +\operatorname {d1} -1} \left (\left (-1\right )^{-\operatorname {a1} -\operatorname {b1} +\operatorname {d1}} \operatorname {hypergeom}\left (\left [\operatorname {d1} , -\operatorname {a1} -\operatorname {b1} +\operatorname {d1} \right ], \left [\operatorname {d1} +1\right ], x\right ) \operatorname {a1} \operatorname {b1} -x^{-\operatorname {d1}} c_{1} \right ) \end {align*}
Integrating both sides gives \begin {align*} y = \int -\left (x -1\right )^{-\operatorname {a1} -\operatorname {b1} +\operatorname {d1} -1} \left (\left (-1\right )^{-\operatorname {a1} -\operatorname {b1} +\operatorname {d1}} \operatorname {hypergeom}\left (\left [\operatorname {d1} , -\operatorname {a1} -\operatorname {b1} +\operatorname {d1} \right ], \left [\operatorname {d1} +1\right ], x\right ) \operatorname {a1} \operatorname {b1} -x^{-\operatorname {d1}} c_{1} \right )d x +c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \int -\left (x -1\right )^{-\operatorname {a1} -\operatorname {b1} +\operatorname {d1} -1} \left (\left (-1\right )^{-\operatorname {a1} -\operatorname {b1} +\operatorname {d1}} \operatorname {hypergeom}\left (\left [\operatorname {d1} , -\operatorname {a1} -\operatorname {b1} +\operatorname {d1} \right ], \left [\operatorname {d1} +1\right ], x\right ) \operatorname {a1} \operatorname {b1} -x^{-\operatorname {d1}} c_{1} \right )d x +c_{2} \\ \end{align*}
Verification of solutions
\[ y = \int -\left (x -1\right )^{-\operatorname {a1} -\operatorname {b1} +\operatorname {d1} -1} \left (\left (-1\right )^{-\operatorname {a1} -\operatorname {b1} +\operatorname {d1}} \operatorname {hypergeom}\left (\left [\operatorname {d1} , -\operatorname {a1} -\operatorname {b1} +\operatorname {d1} \right ], \left [\operatorname {d1} +1\right ], x\right ) \operatorname {a1} \operatorname {b1} -x^{-\operatorname {d1}} c_{1} \right )d x +c_{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}-x \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (\left (\mathit {a1} +\mathit {b1} +1\right ) x -\mathit {d1} \right ) y^{\prime }=-\mathit {a1} \mathit {b1} \mathit {d1} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & \left (x^{2}-x \right ) u^{\prime }\left (x \right )+\left (\left (\mathit {a1} +\mathit {b1} +1\right ) x -\mathit {d1} \right ) u \left (x \right )=-\mathit {a1} \mathit {b1} \mathit {d1} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=-\frac {\left (\left (\mathit {a1} +\mathit {b1} +1\right ) x -\mathit {d1} \right ) u \left (x \right )+\mathit {a1} \mathit {b1} \mathit {d1}}{x^{2}-x} \\ \bullet & {} & \textrm {Collect w.r.t.}\hspace {3pt} u \left (x \right )\hspace {3pt}\textrm {and simplify}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=-\frac {\left (x \mathit {a1} +x \mathit {b1} -\mathit {d1} +x \right ) u \left (x \right )}{x \left (x -1\right )}-\frac {\mathit {a1} \mathit {b1} \mathit {d1}}{x \left (x -1\right )} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} u \left (x \right )\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )+\frac {\left (x \mathit {a1} +x \mathit {b1} -\mathit {d1} +x \right ) u \left (x \right )}{x \left (x -1\right )}=-\frac {\mathit {a1} \mathit {b1} \mathit {d1}}{x \left (x -1\right )} \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (x \right ) \\ {} & {} & \mu \left (x \right ) \left (u^{\prime }\left (x \right )+\frac {\left (x \mathit {a1} +x \mathit {b1} -\mathit {d1} +x \right ) u \left (x \right )}{x \left (x -1\right )}\right )=-\frac {\mu \left (x \right ) \mathit {a1} \mathit {b1} \mathit {d1}}{x \left (x -1\right )} \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d x}\left (u \left (x \right ) \mu \left (x \right )\right ) \\ {} & {} & \mu \left (x \right ) \left (u^{\prime }\left (x \right )+\frac {\left (x \mathit {a1} +x \mathit {b1} -\mathit {d1} +x \right ) u \left (x \right )}{x \left (x -1\right )}\right )=u^{\prime }\left (x \right ) \mu \left (x \right )+u \left (x \right ) \mu ^{\prime }\left (x \right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \mu ^{\prime }\left (x \right ) \\ {} & {} & \mu ^{\prime }\left (x \right )=\frac {\mu \left (x \right ) \left (x \mathit {a1} +x \mathit {b1} -\mathit {d1} +x \right )}{x \left (x -1\right )} \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (x \right )={\mathrm e}^{\ln \left (x -1\right )+\ln \left (x -1\right ) \mathit {a1} +\ln \left (x -1\right ) \mathit {b1} +\mathit {d1} \ln \left (x \right )-\ln \left (x -1\right ) \mathit {d1}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}\left (u \left (x \right ) \mu \left (x \right )\right )\right )d x =\int -\frac {\mu \left (x \right ) \mathit {a1} \mathit {b1} \mathit {d1}}{x \left (x -1\right )}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & u \left (x \right ) \mu \left (x \right )=\int -\frac {\mu \left (x \right ) \mathit {a1} \mathit {b1} \mathit {d1}}{x \left (x -1\right )}d x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {\int -\frac {\mu \left (x \right ) \mathit {a1} \mathit {b1} \mathit {d1}}{x \left (x -1\right )}d x +c_{1}}{\mu \left (x \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (x \right )={\mathrm e}^{\ln \left (x -1\right )+\ln \left (x -1\right ) \mathit {a1} +\ln \left (x -1\right ) \mathit {b1} +\mathit {d1} \ln \left (x \right )-\ln \left (x -1\right ) \mathit {d1}} \\ {} & {} & u \left (x \right )=\frac {\int -\frac {{\mathrm e}^{\ln \left (x -1\right )+\ln \left (x -1\right ) \mathit {a1} +\ln \left (x -1\right ) \mathit {b1} +\mathit {d1} \ln \left (x \right )-\ln \left (x -1\right ) \mathit {d1}} \mathit {a1} \mathit {b1} \mathit {d1}}{x \left (x -1\right )}d x +c_{1}}{{\mathrm e}^{\ln \left (x -1\right )+\ln \left (x -1\right ) \mathit {a1} +\ln \left (x -1\right ) \mathit {b1} +\mathit {d1} \ln \left (x \right )-\ln \left (x -1\right ) \mathit {d1}}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & u \left (x \right )=\left (-\mathit {a1} \mathit {b1} \mathit {d1} \left (\int x^{-1+\mathit {d1}} \left (x -1\right )^{\mathit {a1} +\mathit {b1} -\mathit {d1}}d x \right )+c_{1} \right ) x^{-\mathit {d1}} \left (x -1\right )^{-\mathit {a1} -\mathit {b1} +\mathit {d1} -1} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\left (-\mathit {a1} \mathit {b1} \mathit {d1} \left (\int x^{-1+\mathit {d1}} \left (x -1\right )^{\mathit {a1} +\mathit {b1} -\mathit {d1}}d x \right )+c_{1} \right ) x^{-\mathit {d1}} \left (x -1\right )^{-\mathit {a1} -\mathit {b1} +\mathit {d1} -1} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\left (-\mathit {a1} \mathit {b1} \mathit {d1} \left (\int x^{-1+\mathit {d1}} \left (x -1\right )^{\mathit {a1} +\mathit {b1} -\mathit {d1}}d x \right )+c_{1} \right ) x^{-\mathit {d1}} \left (x -1\right )^{-\mathit {a1} -\mathit {b1} +\mathit {d1} -1} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \left (-\mathit {a1} \mathit {b1} \mathit {d1} \left (\int x^{-1+\mathit {d1}} \left (x -1\right )^{\mathit {a1} +\mathit {b1} -\mathit {d1}}d x \right )+c_{1} \right ) x^{-\mathit {d1}} \left (x -1\right )^{-\mathit {a1} -\mathit {b1} +\mathit {d1} -1}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\int \left (-\mathit {a1} \mathit {b1} \mathrm {signum}\left (x -1\right )^{\mathit {a1} +\mathit {b1} -\mathit {d1}} \left (-\mathrm {signum}\left (x -1\right )\right )^{-\mathit {a1} -\mathit {b1} +\mathit {d1}} x^{\mathit {d1}} \mathit {hypergeom}\left (\left [\mathit {d1} , -\mathit {a1} -\mathit {b1} +\mathit {d1} \right ], \left [\mathit {d1} +1\right ], x\right )+c_{1} \right ) x^{-\mathit {d1}} \left (x -1\right )^{-\mathit {a1} -\mathit {b1} +\mathit {d1} -1}d x +c_{2} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable -> Calling odsolve with the ODE`, diff(_b(_a), _a) = -(_b(_a)*_a*a1+_b(_a)*_a*b1+a1*b1*d1+_b(_a)*_a-_b(_a)*d1)/(_a*(_a-1)), _b(_a)` Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful <- high order exact linear fully integrable successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 76
dsolve(x*(x-1)*diff(diff(y(x),x),x)+((a1+b1+1)*x-d1)*diff(y(x),x)+a1*b1*d1=0,y(x), singsol=all)
\[ y \left (x \right ) = \int \left (x -1\right )^{-\operatorname {a1} -\operatorname {b1} -1+\operatorname {d1}} \left (-\operatorname {a1} \operatorname {b1} \operatorname {signum}\left (x -1\right )^{\operatorname {a1} +\operatorname {b1} -\operatorname {d1}} \left (-\operatorname {signum}\left (x -1\right )\right )^{-\operatorname {a1} -\operatorname {b1} +\operatorname {d1}} \operatorname {hypergeom}\left (\left [\operatorname {d1} , -\operatorname {a1} -\operatorname {b1} +\operatorname {d1} \right ], \left [1+\operatorname {d1} \right ], x\right )+x^{-\operatorname {d1}} c_{1} \right )d x +c_{2} \]
✓ Solution by Mathematica
Time used: 0.618 (sec). Leaf size: 65
DSolve[a1*b1*d1 + (-d1 + (1 + a1 + b1)*x)*y'[x] + (-1 + x)*x*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \text {a1} \text {b1} x \operatorname {Gamma}(\text {d1}+1) \, _3\tilde {F}_2(1,\text {a1}+\text {b1}+1,1;\text {d1}+1,2;x)-\frac {c_1 x^{1-\text {d1}} \operatorname {Hypergeometric2F1}(1-\text {d1},\text {a1}+\text {b1}-\text {d1}+1,2-\text {d1},x)}{\text {d1}-1}+c_2 \]