Internal problem ID [3887]
Internal file name [OUTPUT/3380_Sunday_June_05_2022_09_14_47_AM_27272874/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 23
Problem number: 640.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "homogeneousTypeD2"
Maple gives the following as the ode type
[[_homogeneous, `class D`], _rational]
\[ \boxed {x \left (x^{2}+y^{2}\right ) y^{\prime }-\left (x^{2}+x^{4}+y^{2}\right ) y=0} \]
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 (x^{2}+u \left (x \right )^{2} x^{2}\right ) \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right )-\left (x^{2}+x^{4}+u \left (x \right )^{2} x^{2}\right ) u \left (x \right ) x = 0 \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {u x}{u^{2}+1} \end {align*}
Where \(f(x)=x\) and \(g(u)=\frac {u}{u^{2}+1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u}{u^{2}+1}} \,du &= x \,d x \\ \int { \frac {1}{\frac {u}{u^{2}+1}} \,du} &= \int {x \,d x} \\ \frac {u^{2}}{2}+\ln \left (u \right )&=\frac {x^{2}}{2}+c_{2} \\ \end{align*} The solution is \[ \frac {u \left (x \right )^{2}}{2}+\ln \left (u \left (x \right )\right )-\frac {x^{2}}{2}-c_{2} = 0 \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \frac {y^{2}}{2 x^{2}}+\ln \left (\frac {y}{x}\right )-\frac {x^{2}}{2}-c_{2} = 0\\ \frac {y^{2}}{2 x^{2}}+\ln \left (\frac {y}{x}\right )-\frac {x^{2}}{2}-c_{2} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{2 x^{2}}+\ln \left (\frac {y}{x}\right )-\frac {x^{2}}{2}-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {y^{2}}{2 x^{2}}+\ln \left (\frac {y}{x}\right )-\frac {x^{2}}{2}-c_{2} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (x^{2}+y^{2}\right ) y^{\prime }-\left (x^{2}+x^{4}+y^{2}\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\left (x^{2}+x^{4}+y^{2}\right ) y}{x \left (x^{2}+y^{2}\right )} \end {array} \]
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: 0.031 (sec). Leaf size: 36
dsolve(x*(x^2+y(x)^2)*diff(y(x),x) = (x^2+x^4+y(x)^2)*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{\frac {x^{2}}{2}+c_{1}} x}{\sqrt {\frac {{\mathrm e}^{x^{2}+2 c_{1}}}{\operatorname {LambertW}\left ({\mathrm e}^{x^{2}+2 c_{1}}\right )}}} \]
✓ Solution by Mathematica
Time used: 5.133 (sec). Leaf size: 49
DSolve[x(x^2+y[x]^2)y'[x]==(x^2+x^4+y[x]^2)y[x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -x \sqrt {W\left (e^{x^2+2 c_1}\right )} \\ y(x)\to x \sqrt {W\left (e^{x^2+2 c_1}\right )} \\ y(x)\to 0 \\ \end{align*}