Internal problem ID [3598]
Internal file name [OUTPUT/3091_Sunday_June_05_2022_08_51_50_AM_97358758/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 12
Problem number: 342.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "abelFirstKind"
Maple gives the following as the ode type
[_rational, _Abel]
Unable to solve or complete the solution.
\[ \boxed {\left (b x +a \right )^{2} y^{\prime }+c y^{2}+\left (b x +a \right ) y^{3}=0} \]
This is Abel first kind ODE, it has the form \[ y^{\prime }= f_0(x)+f_1(x) y +f_2(x)y^{2}+f_3(x)y^{3} \] Comparing the above to given ODE which is \begin {align*} y^{\prime }&=-\frac {y^{3}}{b x +a}-\frac {c y^{2}}{\left (b x +a \right )^{2}}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= 0\\ f_1(x) &= 0\\ f_2(x) &= -\frac {c}{\left (b x +a \right )^{2}}\\ f_3(x) &= -\frac {1}{b x +a} \end {align*}
Since \(f_2(x)=-\frac {c}{\left (b x +a \right )^{2}}\) is not zero, then the first step is to apply the following transformation to remove \(f_2\). Let \(y = u(x) - \frac {f_2}{3 f_3}\) or \begin {align*} y &= u(x) - \left ( \frac {-\frac {c}{\left (b x +a \right )^{2}}}{-\frac {3}{b x +a}} \right ) \\ &= u \left (x \right )-\frac {c}{3 b x +3 a} \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = -\frac {u \left (x \right )^{3} b^{3} x^{3}}{\left (b x +a \right )^{4}}-\frac {3 u \left (x \right )^{3} a \,b^{2} x^{2}}{\left (b x +a \right )^{4}}-\frac {3 u \left (x \right )^{3} a^{2} b x}{\left (b x +a \right )^{4}}-\frac {u \left (x \right )^{3} a^{3}}{\left (b x +a \right )^{4}}-\frac {b^{3} c \,x^{2}}{3 \left (b x +a \right )^{4}}+\frac {u \left (x \right ) b \,c^{2} x}{3 \left (b x +a \right )^{4}}-\frac {2 a \,b^{2} c x}{3 \left (b x +a \right )^{4}}+\frac {u \left (x \right ) a \,c^{2}}{3 \left (b x +a \right )^{4}}-\frac {a^{2} b c}{3 \left (b x +a \right )^{4}}-\frac {2 c^{3}}{27 \left (b x +a \right )^{4}}\tag {2} \end {align*}
This is Abel first kind ODE, it has the form \[ u^{\prime }\left (x \right )= f_0(x)+f_1(x) u \left (x \right ) +f_2(x)u \left (x \right )^{2}+f_3(x)u \left (x \right )^{3} \] Comparing the above to given ODE which is \begin {align*} u^{\prime }\left (x \right )&=-\frac {\left (27 b^{3} x^{3}+81 a \,b^{2} x^{2}+81 a^{2} b x +27 a^{3}\right ) u \left (x \right )^{3}}{27 \left (b^{4} x^{4}+4 x^{3} b^{3} a +6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}\right )}-\frac {\left (-9 b \,c^{2} x -9 c^{2} a \right ) u \left (x \right )}{27 \left (b^{4} x^{4}+4 x^{3} b^{3} a +6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}\right )}-\frac {9 b^{3} c \,x^{2}+18 a \,b^{2} c x +9 a^{2} b c +2 c^{3}}{27 \left (b^{4} x^{4}+4 x^{3} b^{3} a +6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}\right )}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= -\frac {b^{3} c \,x^{2}}{3 \left (b^{4} x^{4}+4 x^{3} b^{3} a +6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}\right )}-\frac {2 a \,b^{2} c x}{3 \left (b^{4} x^{4}+4 x^{3} b^{3} a +6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}\right )}-\frac {a^{2} b c}{3 \left (b^{4} x^{4}+4 x^{3} b^{3} a +6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}\right )}-\frac {2 c^{3}}{27 \left (b^{4} x^{4}+4 x^{3} b^{3} a +6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}\right )}\\ f_1(x) &= \frac {b \,c^{2} x}{3 b^{4} x^{4}+12 x^{3} b^{3} a +18 a^{2} b^{2} x^{2}+12 a^{3} b x +3 a^{4}}+\frac {c^{2} a}{3 b^{4} x^{4}+12 x^{3} b^{3} a +18 a^{2} b^{2} x^{2}+12 a^{3} b x +3 a^{4}}\\ f_2(x) &= 0\\ f_3(x) &= -\frac {b^{3} x^{3}}{b^{4} x^{4}+4 x^{3} b^{3} a +6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}}-\frac {3 a \,b^{2} x^{2}}{b^{4} x^{4}+4 x^{3} b^{3} a +6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}}-\frac {3 a^{2} b x}{b^{4} x^{4}+4 x^{3} b^{3} a +6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}}-\frac {a^{3}}{b^{4} x^{4}+4 x^{3} b^{3} a +6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}} \end {align*}
Since \(f_2(x)=0\) then we check the Abel invariant to see if it depends on \(x\) or not. The Abel invariant is given by \begin {align*} -\frac {f_{1}^{3}}{f_{0}^{2} f_{3}} \end {align*}
Which when evaluating gives \begin {align*} \text {Expression too large to display} \end {align*}
Since the Abel invariant depends on \(x\) then unable to solve this ode at this time.
Unable to complete the solution now.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (b x +a \right )^{2} y^{\prime }+c y^{2}+\left (b x +a \right ) y^{3}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {c y^{2}+\left (b x +a \right ) y^{3}}{\left (b x +a \right )^{2}} \end {array} \]
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 differential order: 1; looking for linear symmetries trying exact trying Abel <- Abel successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 126
dsolve((b*x+a)^2*diff(y(x),x)+c*y(x)^2+(b*x+a)*y(x)^3 = 0,y(x), singsol=all)
\[ \frac {\left (\sqrt {b}\, a +b^{\frac {3}{2}} x \right ) {\mathrm e}^{-\frac {\left (\left (b x +a +c \right ) y \left (x \right )+b \left (b x +a \right )\right ) \left (\left (-b x -a +c \right ) y \left (x \right )+b \left (b x +a \right )\right )}{2 y \left (x \right )^{2} \left (b x +a \right )^{2} b}}+\frac {c \sqrt {2}\, \sqrt {\pi }\, {\mathrm e}^{\frac {1}{2 b}} \operatorname {erf}\left (\frac {\sqrt {2}\, \left (c y \left (x \right )+b \left (b x +a \right )\right )}{2 \sqrt {b}\, y \left (x \right ) \left (b x +a \right )}\right )}{2}+b^{\frac {3}{2}} c_{1}}{b^{\frac {3}{2}}} = 0 \]
✓ Solution by Mathematica
Time used: 1.435 (sec). Leaf size: 149
DSolve[(a+b x)^2 y'[x]+c y[x]^2+(a+b x)y[x]^3==0,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [-\frac {c}{\sqrt {-b (a+b x)^2}}=\frac {2 \exp \left (\frac {1}{2} \left (-\frac {c}{\sqrt {-b (a+b x)^2}}-\frac {\left (-b (a+b x)^2\right )^{3/2}}{b y(x) (a+b x)^3}\right )^2\right )}{\sqrt {2 \pi } \text {erfi}\left (\frac {-\frac {c}{\sqrt {-b (a+b x)^2}}-\frac {\left (-b (a+b x)^2\right )^{3/2}}{b y(x) (a+b x)^3}}{\sqrt {2}}\right )+2 c_1},y(x)\right ] \]