73.6.19 problem 7.5 (i)

Internal problem ID [15158]
Book : Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section : Chapter 7. The exact form and general integrating fators. Additional exercises. page 141
Problem number : 7.5 (i)
Date solved : Tuesday, January 28, 2025 at 07:38:44 AM
CAS classification : [[_homogeneous, `class G`], _rational]

\begin{align*} 6+12 x^{2} y^{2}+\left (7 x^{3} y+\frac {x}{y}\right ) y^{\prime }&=0 \end{align*}

✓ Solution by Maple

Time used: 2.260 (sec). Leaf size: 63

dsolve(6+12*x^2*y(x)^2+(7*x^3*y(x)+x/y(x))*diff(y(x),x)=0,y(x), singsol=all)
 

\[ y = \operatorname {RootOf}\left (\textit {\_Z}^{35} c_{1}^{2} x^{10}-\textit {\_Z}^{30} c_{1}^{2} x^{10}-1\right )^{15} x^{4} \left (\operatorname {RootOf}\left (\textit {\_Z}^{35} c_{1}^{2} x^{10}-\textit {\_Z}^{30} c_{1}^{2} x^{10}-1\right )^{5}-1\right ) c_{1} \]

✓ Solution by Mathematica

Time used: 0.209 (sec). Leaf size: 399

DSolve[6+12*x^2*y[x]^2+(7*x^3*y[x]+x/y[x])*D[y[x],x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ \text {Solve}\left [\int _1^{y(x)}\left (7 \exp \left (\int _1^{x K[3]}-\frac {3 K[1]^2+1}{K[1]^3+K[1]}dK[1]\right ) K[3]^2 x^3+\exp \left (\int _1^{x K[3]}-\frac {3 K[1]^2+1}{K[1]^3+K[1]}dK[1]\right ) x-\int _1^x\left (-\frac {12 \exp \left (\int _1^{K[2] K[3]}-\frac {3 K[1]^2+1}{K[1]^3+K[1]}dK[1]\right ) K[2]^3 \left (3 K[2]^2 K[3]^2+1\right ) K[3]^3}{K[2]^3 K[3]^3+K[2] K[3]}+36 \exp \left (\int _1^{K[2] K[3]}-\frac {3 K[1]^2+1}{K[1]^3+K[1]}dK[1]\right ) K[2]^2 K[3]^2-\frac {6 \exp \left (\int _1^{K[2] K[3]}-\frac {3 K[1]^2+1}{K[1]^3+K[1]}dK[1]\right ) K[2] \left (3 K[2]^2 K[3]^2+1\right ) K[3]}{K[2]^3 K[3]^3+K[2] K[3]}+6 \exp \left (\int _1^{K[2] K[3]}-\frac {3 K[1]^2+1}{K[1]^3+K[1]}dK[1]\right )\right )dK[2]\right )dK[3]+\int _1^x\left (12 \exp \left (\int _1^{K[2] y(x)}-\frac {3 K[1]^2+1}{K[1]^3+K[1]}dK[1]\right ) K[2]^2 y(x)^3+6 \exp \left (\int _1^{K[2] y(x)}-\frac {3 K[1]^2+1}{K[1]^3+K[1]}dK[1]\right ) y(x)\right )dK[2]=c_1,y(x)\right ] \]