60.1.450 problem 453
Internal
problem
ID
[10464]
Book
:
Differential
Gleichungen,
E.
Kamke,
3rd
ed.
Chelsea
Pub.
NY,
1948
Section
:
Chapter
1,
linear
first
order
Problem
number
:
453
Date
solved
:
Monday, January 27, 2025 at 07:47:52 PM
CAS
classification
:
[[_homogeneous, `class A`], _dAlembert]
\begin{align*} \left (a^{2}-1\right ) x^{2} {y^{\prime }}^{2}+2 x y y^{\prime }-y^{2}+a^{2} x^{2}&=0 \end{align*}
✓ Solution by Maple
Time used: 0.074 (sec). Leaf size: 229
dsolve((a^2-1)*x^2*diff(y(x),x)^2+2*x*y(x)*diff(y(x),x)-y(x)^2+a^2*x^2 = 0,y(x), singsol=all)
\begin{align*}
\frac {2 a \ln \left (x \right )-2 \sqrt {-a^{2}}\, \arctan \left (\frac {a^{2} y}{\sqrt {-a^{2}}\, \sqrt {\frac {-a^{2} x^{2}+x^{2}+y^{2}}{x^{2}}}\, x}\right )+\ln \left (\frac {x^{2}+y^{2}}{x^{2}}\right ) a -2 c_{1} a +2 \ln \left (\frac {\sqrt {\frac {-a^{2} x^{2}+x^{2}+y^{2}}{x^{2}}}\, x +y}{x}\right )}{2 a} &= 0 \\
\frac {2 a \ln \left (x \right )+2 \sqrt {-a^{2}}\, \arctan \left (\frac {a^{2} y}{\sqrt {-a^{2}}\, \sqrt {\frac {-a^{2} x^{2}+x^{2}+y^{2}}{x^{2}}}\, x}\right )+\ln \left (\frac {x^{2}+y^{2}}{x^{2}}\right ) a -2 c_{1} a -2 \ln \left (\frac {\sqrt {\frac {-a^{2} x^{2}+x^{2}+y^{2}}{x^{2}}}\, x +y}{x}\right )}{2 a} &= 0 \\
\end{align*}
✓ Solution by Mathematica
Time used: 1.017 (sec). Leaf size: 223
DSolve[a^2*x^2 - y[x]^2 + 2*x*y[x]*D[y[x],x] + (-1 + a^2)*x^2*D[y[x],x]^2==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*}
\text {Solve}\left [\frac {2 i \arctan \left (\frac {y(x)}{x \sqrt {a^2-\frac {y(x)^2}{x^2}-1}}\right )-2 i a \arctan \left (\frac {a y(x)}{x \sqrt {a^2-\frac {y(x)^2}{x^2}-1}}\right )+a \log \left (\frac {y(x)^2}{x^2}+1\right )}{2 a^2-2}&=\frac {a \log \left (x-a^2 x\right )}{1-a^2}+c_1,y(x)\right ] \\
\text {Solve}\left [\frac {-2 i \arctan \left (\frac {y(x)}{x \sqrt {a^2-\frac {y(x)^2}{x^2}-1}}\right )+2 i a \arctan \left (\frac {a y(x)}{x \sqrt {a^2-\frac {y(x)^2}{x^2}-1}}\right )+a \log \left (\frac {y(x)^2}{x^2}+1\right )}{2 a^2-2}&=\frac {a \log \left (x-a^2 x\right )}{1-a^2}+c_1,y(x)\right ] \\
\end{align*}