60.7.97 problem 1688 (book 6.97)

Internal problem ID [11686]
Book : Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section : Chapter 6, non-linear second order
Problem number : 1688 (book 6.97)
Date solved : Tuesday, January 28, 2025 at 06:07:25 PM
CAS classification : [[_2nd_order, _with_linear_symmetries], [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]

\begin{align*} x^{4} y^{\prime \prime }-x \left (x^{2}+2 y\right ) y^{\prime }+4 y^{2}&=0 \end{align*}

✓ Solution by Maple

Time used: 0.058 (sec). Leaf size: 22

dsolve(x^4*diff(diff(y(x),x),x)-x*(x^2+2*y(x))*diff(y(x),x)+4*y(x)^2=0,y(x), singsol=all)
 

\[ y = \left (-\tanh \left (c_{1} \left (\ln \left (x \right )-c_{2} \right )\right ) c_{1} +1\right ) x^{2} \]

✓ Solution by Mathematica

Time used: 0.231 (sec). Leaf size: 178

DSolve[4*y[x]^2 - x*(x^2 + 2*y[x])*D[y[x],x] + x^4*D[y[x],{x,2}] == 0,y[x],x,IncludeSingularSolutions -> True]
 

\[ \text {Solve}\left [\int _1^{y(x)}\frac {x^2}{-c_1 x^4+2 K[2] x^2-K[2]^2}dK[2]-\int _1^x\left (\frac {\left (\frac {y(x)^2}{K[3]^4}+c_1\right ) K[3]^3}{-c_1 K[3]^4+2 y(x) K[3]^2-y(x)^2}+\int _1^{y(x)}\left (\frac {2 K[3]}{-c_1 K[3]^4+2 K[2] K[3]^2-K[2]^2}-\frac {K[3]^2 \left (4 K[2] K[3]-4 c_1 K[3]^3\right )}{\left (-c_1 K[3]^4+2 K[2] K[3]^2-K[2]^2\right ){}^2}\right )dK[2]\right )dK[3]=c_2,y(x)\right ] \]