Internal problem ID [4208]
Internal file name [OUTPUT/3701_Sunday_June_05_2022_10_18_09_AM_33411510/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 33
Problem number: 975.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "exact", "quadrature", "separable", "differentialType", "homogeneousTypeD2", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[_quadrature]
\[ \boxed {y^{2} {y^{\prime }}^{2}-\left (x +1\right ) y y^{\prime }=-x} \] The ode \begin {align*} y^{2} {y^{\prime }}^{2}-\left (x +1\right ) y y^{\prime } = -x \end {align*}
is factored to \begin {align*} \left (y^{\prime } y-1\right ) \left (y^{\prime } y-x \right ) = 0 \end {align*}
Which gives the following equations \begin {align*} y^{\prime } y-1 = 0\tag {1} \\ y^{\prime } y-x = 0\tag {2} \\ \end {align*}
Each of the above equations is now solved.
Solving ODE (1) Integrating both sides gives \begin {align*} \int y d y &= x +c_{1}\\ \frac {y^{2}}{2}&=x +c_{1} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\sqrt {2 x +2 c_{1}}\\ y_2&=-\sqrt {2 x +2 c_{1}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {2 x +2 c_{1}} \\ \tag{2} y &= -\sqrt {2 x +2 c_{1}} \\ \end{align*}
Verification of solutions
\[ y = \sqrt {2 x +2 c_{1}} \] Verified OK.
\[ y = -\sqrt {2 x +2 c_{1}} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {2 x +2 c_{1}} \\ \tag{2} y &= -\sqrt {2 x +2 c_{1}} \\ \end{align*}
Verification of solutions
\[ y = \sqrt {2 x +2 c_{1}} \] Verified OK.
\[ y = -\sqrt {2 x +2 c_{1}} \] 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.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{2} {y^{\prime }}^{2}-\left (x +1\right ) y y^{\prime }=-x \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\frac {1}{y}, y^{\prime }=\frac {x}{y}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {1}{y} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & y^{\prime } y=1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime } yd x =\int 1d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {y^{2}}{2}=x +c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\sqrt {2 x +2 c_{1}}, y=-\sqrt {2 x +2 c_{1}}\right \} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {x}{y} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & y^{\prime } y=x \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime } yd x =\int x d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {y^{2}}{2}=\frac {x^{2}}{2}+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\sqrt {x^{2}+2 c_{1}}, y=-\sqrt {x^{2}+2 c_{1}}\right \} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\left \{y=\sqrt {2 x +2 c_{1}}, y=-\sqrt {2 x +2 c_{1}}\right \}, \left \{y=\sqrt {x^{2}+2 c_{1}}, y=-\sqrt {x^{2}+2 c_{1}}\right \}\right \} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli <- Bernoulli 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: 45
dsolve(y(x)^2*diff(y(x),x)^2-(1+x)*y(x)*diff(y(x),x)+x = 0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \sqrt {2 x +c_{1}} \\ y \left (x \right ) &= -\sqrt {2 x +c_{1}} \\ 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: 0.152 (sec). Leaf size: 72
DSolve[y[x]^2 (y'[x])^2-(1+x)y[x] y'[x]+x==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\sqrt {2} \sqrt {x+c_1} \\ y(x)\to \sqrt {2} \sqrt {x+c_1} \\ y(x)\to -\sqrt {x^2+2 c_1} \\ y(x)\to \sqrt {x^2+2 c_1} \\ \end{align*}