Internal problem ID [8537]
Internal file name [OUTPUT/7470_Sunday_June_05_2022_10_57_23_PM_87655422/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 201.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[_Riccati]
\[ \boxed {2 f \left (x \right ) y^{\prime }+2 f \left (x \right ) y^{2}-f^{\prime }\left (x \right ) y=2 f \left (x \right )^{2}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {-2 f \left (x \right ) y^{2}+f^{\prime }\left (x \right ) y +2 f \left (x \right )^{2}}{2 f \left (x \right )} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -y^{2}+\frac {f^{\prime }\left (x \right ) y}{2 f \left (x \right )}+f \left (x \right ) \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=f \left (x \right )\), \(f_1(x)=\frac {f^{\prime }\left (x \right )}{2 f \left (x \right )}\) and \(f_2(x)=-1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-u} \tag {1} \end {align*}
Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for \(u(x)\) which is \begin {align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end {align*}
But \begin {align*} f_2' &=0\\ f_1 f_2 &=-\frac {f^{\prime }\left (x \right )}{2 f \left (x \right )}\\ f_2^2 f_0 &=f \left (x \right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -u^{\prime \prime }\left (x \right )+\frac {f^{\prime }\left (x \right ) u^{\prime }\left (x \right )}{2 f \left (x \right )}+f \left (x \right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} {\mathrm e}^{i \left (\int \sqrt {-f \left (x \right )}d x \right )}+c_{2} {\mathrm e}^{-i \left (\int \sqrt {-f \left (x \right )}d x \right )} \] The above shows that \[ u^{\prime }\left (x \right ) = i \sqrt {-f \left (x \right )}\, \left (c_{1} {\mathrm e}^{i \left (\int \sqrt {-f \left (x \right )}d x \right )}-c_{2} {\mathrm e}^{-i \left (\int \sqrt {-f \left (x \right )}d x \right )}\right ) \] Using the above in (1) gives the solution \[ y = \frac {i \sqrt {-f \left (x \right )}\, \left (c_{1} {\mathrm e}^{i \left (\int \sqrt {-f \left (x \right )}d x \right )}-c_{2} {\mathrm e}^{-i \left (\int \sqrt {-f \left (x \right )}d x \right )}\right )}{c_{1} {\mathrm e}^{i \left (\int \sqrt {-f \left (x \right )}d x \right )}+c_{2} {\mathrm e}^{-i \left (\int \sqrt {-f \left (x \right )}d x \right )}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {i \sqrt {-f \left (x \right )}\, \left (c_{1} {\mathrm e}^{i \left (\int \sqrt {-f \left (x \right )}d x \right )}-c_{2} {\mathrm e}^{-i \left (\int \sqrt {-f \left (x \right )}d x \right )}\right )}{c_{1} {\mathrm e}^{i \left (\int \sqrt {-f \left (x \right )}d x \right )}+c_{2} {\mathrm e}^{-i \left (\int \sqrt {-f \left (x \right )}d x \right )}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {i \sqrt {-f \left (x \right )}\, \left (c_{1} {\mathrm e}^{i \left (\int \sqrt {-f \left (x \right )}d x \right )}-c_{2} {\mathrm e}^{-i \left (\int \sqrt {-f \left (x \right )}d x \right )}\right )}{c_{1} {\mathrm e}^{i \left (\int \sqrt {-f \left (x \right )}d x \right )}+c_{2} {\mathrm e}^{-i \left (\int \sqrt {-f \left (x \right )}d x \right )}} \\ \end{align*}
Verification of solutions
\[ y = \frac {i \sqrt {-f \left (x \right )}\, \left (c_{1} {\mathrm e}^{i \left (\int \sqrt {-f \left (x \right )}d x \right )}-c_{2} {\mathrm e}^{-i \left (\int \sqrt {-f \left (x \right )}d x \right )}\right )}{c_{1} {\mathrm e}^{i \left (\int \sqrt {-f \left (x \right )}d x \right )}+c_{2} {\mathrm e}^{-i \left (\int \sqrt {-f \left (x \right )}d x \right )}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 f \left (x \right ) y^{\prime }+2 f \left (x \right ) y^{2}-f^{\prime }\left (x \right ) y=2 f \left (x \right )^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-2 f \left (x \right ) y^{2}+f^{\prime }\left (x \right ) y+2 f \left (x \right )^{2}}{2 f \left (x \right )} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying Chini <- Chini successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 23
dsolve(2*f(x)*diff(y(x),x)+2*f(x)*y(x)^2-diff(f(x),x)*y(x)-2*f(x)^2=0,y(x), singsol=all)
\[ y \left (x \right ) = i \tan \left (-i \left (\int \sqrt {f \left (x \right )}d x \right )+c_{1} \right ) \sqrt {f \left (x \right )} \]
✓ Solution by Mathematica
Time used: 0.256 (sec). Leaf size: 39
DSolve[2*f[x]*y'[x]+2*f[x]*y[x]^2-f'[x]*y[x]-2*f[x]^2==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to i \sqrt {f(x)} \tan \left (i \int _1^x-\sqrt {f(K[1])}dK[1]+c_1\right ) \]