Internal problem ID [4195]
Internal file name [OUTPUT/3688_Sunday_June_05_2022_10_12_56_AM_9699418/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 32
Problem number: 962.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "exact", "linear", "separable", "differentialType", "homogeneousTypeD2", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[_separable]
\[ \boxed {x y {y^{\prime }}^{2}+\left (x^{2}+y^{2}\right ) y^{\prime }+y x=0} \] The ode \begin {align*} x y {y^{\prime }}^{2}+\left (x^{2}+y^{2}\right ) y^{\prime }+y x = 0 \end {align*}
is factored to \begin {align*} \left (x y^{\prime }+y\right ) \left (y^{\prime } y+x \right ) = 0 \end {align*}
Which gives the following equations \begin {align*} x y^{\prime }+y = 0\tag {1} \\ y^{\prime } y+x = 0\tag {2} \\ \end {align*}
Each of the above equations is now solved.
Solving ODE (1) In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= -\frac {y}{x} \end {align*}
Where \(f(x)=-\frac {1}{x}\) and \(g(y)=y\). Integrating both sides gives \begin {align*} \frac {1}{y} \,dy &= -\frac {1}{x} \,d x\\ \int { \frac {1}{y} \,dy} &= \int {-\frac {1}{x} \,d x}\\ \ln \left (y \right )&=-\ln \left (x \right )+c_{1}\\ y&={\mathrm e}^{-\ln \left (x \right )+c_{1}}\\ &=\frac {c_{1}}{x} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1}}{x} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1}}{x} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1}}{x} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1}}{x} \] Verified OK.
Solving ODE (2) In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= -\frac {x}{y} \end {align*}
Where \(f(x)=-x\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= -x \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {-x \,d x} \\ \frac {y^{2}}{2}&=-\frac {x^{2}}{2}+c_{2} \\ \end{align*} Which results in \begin{align*} y &= \sqrt {-x^{2}+2 c_{2}} \\ y &= -\sqrt {-x^{2}+2 c_{2}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {-x^{2}+2 c_{2}} \\ \tag{2} y &= -\sqrt {-x^{2}+2 c_{2}} \\ \end{align*}
Verification of solutions
\[ y = \sqrt {-x^{2}+2 c_{2}} \] Verified OK.
\[ y = -\sqrt {-x^{2}+2 c_{2}} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {-x^{2}+2 c_{2}} \\ \tag{2} y &= -\sqrt {-x^{2}+2 c_{2}} \\ \end{align*}
Verification of solutions
\[ y = \sqrt {-x^{2}+2 c_{2}} \] Verified OK.
\[ y = -\sqrt {-x^{2}+2 c_{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y {y^{\prime }}^{2}+\left (x^{2}+y^{2}\right ) y^{\prime }+y x =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=-\frac {x}{y}, y^{\prime }=-\frac {y}{x}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {x}{y} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & y^{\prime } y=-x \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime } yd x =\int -x d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {y^{2}}{2}=-\frac {x^{2}}{2}+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\sqrt {-x^{2}+2 c_{1}}, y=-\sqrt {-x^{2}+2 c_{1}}\right \} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {y}{x} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=-\frac {1}{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int -\frac {1}{x}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=-\ln \left (x \right )+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {{\mathrm e}^{c_{1}}}{x} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y=\frac {{\mathrm e}^{c_{1}}}{x}, \left \{y=\sqrt {-x^{2}+2 c_{1}}, y=-\sqrt {-x^{2}+2 c_{1}}\right \}\right \} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli <- Bernoulli successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 35
dsolve(x*y(x)*diff(y(x),x)^2+(x^2+y(x)^2)*diff(y(x),x)+x*y(x) = 0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {c_{1}}{x} \\ y \left (x \right ) &= \sqrt {-x^{2}+c_{1}} \\ y \left (x \right ) &= -\sqrt {-x^{2}+c_{1}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.142 (sec). Leaf size: 54
DSolve[x y[x] (y'[x])^2+(x^2 + y[x]^2)y'[x]+x y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {c_1}{x} \\ y(x)\to -\sqrt {-x^2+2 c_1} \\ y(x)\to \sqrt {-x^2+2 c_1} \\ y(x)\to 0 \\ \end{align*}