Internal problem ID [3606]
Internal file name [OUTPUT/3099_Sunday_June_05_2022_08_52_05_AM_97534647/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 12
Problem number: 350.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[_rational, _Riccati]
\[ \boxed {y^{\prime } x^{3}+\left (-2 x +3\right ) x^{2} y-x^{6} y^{2}=-3} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {x^{6} y^{2}+2 x^{3} y -3 x^{2} y -3}{x^{3}} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2} x^{3}+2 y -\frac {3 y}{x}-\frac {3}{x^{3}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-\frac {3}{x^{3}}\), \(f_1(x)=\frac {2 x^{3}-3 x^{2}}{x^{3}}\) and \(f_2(x)=x^{3}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{x^{3} 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' &=3 x^{2}\\ f_1 f_2 &=2 x^{3}-3 x^{2}\\ f_2^2 f_0 &=-3 x^{3} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} x^{3} u^{\prime \prime }\left (x \right )-2 x^{3} u^{\prime }\left (x \right )-3 x^{3} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} {\mathrm e}^{3 x}+c_{2} {\mathrm e}^{-x} \] The above shows that \[ u^{\prime }\left (x \right ) = 3 c_{1} {\mathrm e}^{3 x}-c_{2} {\mathrm e}^{-x} \] Using the above in (1) gives the solution \[ y = -\frac {3 c_{1} {\mathrm e}^{3 x}-c_{2} {\mathrm e}^{-x}}{x^{3} \left (c_{1} {\mathrm e}^{3 x}+c_{2} {\mathrm e}^{-x}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {-3 c_{3} {\mathrm e}^{4 x}+1}{x^{3} \left (c_{3} {\mathrm e}^{4 x}+1\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-3 c_{3} {\mathrm e}^{4 x}+1}{x^{3} \left (c_{3} {\mathrm e}^{4 x}+1\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {-3 c_{3} {\mathrm e}^{4 x}+1}{x^{3} \left (c_{3} {\mathrm e}^{4 x}+1\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x^{3}+\left (-2 x +3\right ) x^{2} y-x^{6} y^{2}=-3 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-3-\left (-2 x +3\right ) x^{2} y+x^{6} y^{2}}{x^{3}} \end {array} \]
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: 0.0 (sec). Leaf size: 27
dsolve(x^3*diff(y(x),x)+3+(3-2*x)*x^2*y(x)-x^6*y(x)^2 = 0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {-3 \,{\mathrm e}^{4 x} c_{1} -3}{x^{3} \left ({\mathrm e}^{4 x} c_{1} -3\right )} \]
✓ Solution by Mathematica
Time used: 0.173 (sec). Leaf size: 34
DSolve[x^3 y'[x]+3+(3-2 x)x^2 y[x]-x^6 y[x]^2==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {-3+\frac {1}{\frac {1}{4}+c_1 e^{4 x}}}{x^3} \\ y(x)\to -\frac {3}{x^3} \\ \end{align*}