Internal problem ID [3423]
Internal file name [OUTPUT/2916_Sunday_June_05_2022_08_47_01_AM_64069676/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 6
Problem number: 167.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati", "homogeneousTypeD2"
Maple gives the following as the ode type
[[_homogeneous, `class D`], _rational, _Riccati]
\[ \boxed {x y^{\prime }-y-b y^{2}=a \,x^{2}} \]
Using the change of variables \(y = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} x \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right )-u \left (x \right ) x -b u \left (x \right )^{2} x^{2} = a \,x^{2} \end {align*}
Integrating both sides gives \begin {align*} \int \frac {1}{u^{2} b +a}d u &= x +c_{2}\\ \frac {\arctan \left (\frac {b u}{\sqrt {a b}}\right )}{\sqrt {a b}}&=x +c_{2} \end {align*}
Solving for \(u\) gives these solutions \begin {align*} u_1&=\frac {\tan \left (c_{2} \sqrt {a b}+x \sqrt {a b}\right ) \sqrt {a b}}{b} \end {align*}
Therefore the solution \(y\) is \begin {align*} y&=x u\\ &=\frac {x \tan \left (c_{2} \sqrt {a b}+x \sqrt {a b}\right ) \sqrt {a b}}{b} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x \tan \left (c_{2} \sqrt {a b}+x \sqrt {a b}\right ) \sqrt {a b}}{b} \\ \end{align*}
Verification of solutions
\[ y = \frac {x \tan \left (c_{2} \sqrt {a b}+x \sqrt {a b}\right ) \sqrt {a b}}{b} \] Verified OK.
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {a \,x^{2}+b \,y^{2}+y}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = x a +\frac {b \,y^{2}}{x}+\frac {y}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=x a\), \(f_1(x)=\frac {1}{x}\) and \(f_2(x)=\frac {b}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {b u}{x}} \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' &=-\frac {b}{x^{2}}\\ f_1 f_2 &=\frac {b}{x^{2}}\\ f_2^2 f_0 &=\frac {b^{2} a}{x} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {b u^{\prime \prime }\left (x \right )}{x}+\frac {b^{2} a u \left (x \right )}{x} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} \sin \left (\sqrt {a}\, \sqrt {b}\, x \right )+c_{2} \cos \left (\sqrt {a}\, \sqrt {b}\, x \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \sqrt {a}\, \sqrt {b}\, \left (c_{1} \cos \left (\sqrt {a}\, \sqrt {b}\, x \right )-c_{2} \sin \left (\sqrt {a}\, \sqrt {b}\, x \right )\right ) \] Using the above in (1) gives the solution \[ y = -\frac {\sqrt {a}\, \left (c_{1} \cos \left (\sqrt {a}\, \sqrt {b}\, x \right )-c_{2} \sin \left (\sqrt {a}\, \sqrt {b}\, x \right )\right ) x}{\sqrt {b}\, \left (c_{1} \sin \left (\sqrt {a}\, \sqrt {b}\, x \right )+c_{2} \cos \left (\sqrt {a}\, \sqrt {b}\, x \right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (-c_{3} \cos \left (\sqrt {a}\, \sqrt {b}\, x \right )+\sin \left (\sqrt {a}\, \sqrt {b}\, x \right )\right ) \sqrt {a}\, x}{\left (c_{3} \sin \left (\sqrt {a}\, \sqrt {b}\, x \right )+\cos \left (\sqrt {a}\, \sqrt {b}\, x \right )\right ) \sqrt {b}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-c_{3} \cos \left (\sqrt {a}\, \sqrt {b}\, x \right )+\sin \left (\sqrt {a}\, \sqrt {b}\, x \right )\right ) \sqrt {a}\, x}{\left (c_{3} \sin \left (\sqrt {a}\, \sqrt {b}\, x \right )+\cos \left (\sqrt {a}\, \sqrt {b}\, x \right )\right ) \sqrt {b}} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (-c_{3} \cos \left (\sqrt {a}\, \sqrt {b}\, x \right )+\sin \left (\sqrt {a}\, \sqrt {b}\, x \right )\right ) \sqrt {a}\, x}{\left (c_{3} \sin \left (\sqrt {a}\, \sqrt {b}\, x \right )+\cos \left (\sqrt {a}\, \sqrt {b}\, x \right )\right ) \sqrt {b}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y^{\prime }-y-b y^{2}=a \,x^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {a \,x^{2}+y+b y^{2}}{x} \end {array} \]
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 homogeneous D <- homogeneous successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 24
dsolve(x*diff(y(x),x) = a*x^2+y(x)+b*y(x)^2,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\tan \left (\sqrt {a b}\, \left (c_{1} +x \right )\right ) x \sqrt {a b}}{b} \]
✓ Solution by Mathematica
Time used: 17.546 (sec). Leaf size: 33
DSolve[x y'[x]==a x^2+y[x]+b y[x]^2,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {\sqrt {a} x \tan \left (\sqrt {a} \sqrt {b} (x+c_1)\right )}{\sqrt {b}} \]