Internal problem ID [4047]
Internal file name [OUTPUT/3540_Sunday_June_05_2022_09_36_31_AM_89530523/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 28
Problem number: 808.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "exact", "linear", "quadrature", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[_quadrature]
\[ \boxed {{y^{\prime }}^{2}+y^{\prime } y-x \left (y+x \right )=0} \] The ode \begin {align*} {y^{\prime }}^{2}+y^{\prime } y-x \left (y+x \right ) = 0 \end {align*}
is factored to \begin {align*} \left (y^{\prime }-x \right ) \left (y^{\prime }+y+x \right ) = 0 \end {align*}
Which gives the following equations \begin {align*} y^{\prime }-x = 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*} y &= \int { x\,\mathop {\mathrm {d}x}}\\ &= \frac {x^{2}}{2}+c_{1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{2}}{2}+c_{1} \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{2}}{2}+c_{1} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{2}}{2}+c_{1} \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{2}}{2}+c_{1} \] Verified OK.
Solving ODE (2)
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=1\\ q(x) &=-x \end {align*}
Hence the ode is \begin {align*} y^{\prime }+y = -x \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int 1d x} \\ &= {\mathrm e}^{x} \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (-x\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (y \,{\mathrm e}^{x}\right ) &= \left ({\mathrm e}^{x}\right ) \left (-x\right )\\ \mathrm {d} \left (y \,{\mathrm e}^{x}\right ) &= \left (-x \,{\mathrm e}^{x}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} y \,{\mathrm e}^{x} &= \int {-x \,{\mathrm e}^{x}\,\mathrm {d} x}\\ y \,{\mathrm e}^{x} &= -\left (-1+x \right ) {\mathrm e}^{x} + c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{x}\) results in \begin {align*} y &= -{\mathrm e}^{-x} \left (-1+x \right ) {\mathrm e}^{x}+c_{2} {\mathrm e}^{-x} \end {align*}
which simplifies to \begin {align*} y &= 1-x +c_{2} {\mathrm e}^{-x} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= 1-x +c_{2} {\mathrm e}^{-x} \\ \end{align*}
Verification of solutions
\[ y = 1-x +c_{2} {\mathrm e}^{-x} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= 1-x +c_{2} {\mathrm e}^{-x} \\ \end{align*}
Verification of solutions
\[ y = 1-x +c_{2} {\mathrm e}^{-x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & {y^{\prime }}^{2}+y^{\prime } y-x \left (y+x \right )=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 }=x , y^{\prime }=-x -y\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=x \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int x d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=\frac {x^{2}}{2}+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {x^{2}}{2}+c_{1} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-x -y \\ {} & \circ & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & y^{\prime }+y=-x \\ {} & \circ & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (x \right ) \\ {} & {} & \mu \left (x \right ) \left (y^{\prime }+y\right )=-\mu \left (x \right ) x \\ {} & \circ & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d x}\left (y \mu \left (x \right )\right ) \\ {} & {} & \mu \left (x \right ) \left (y^{\prime }+y\right )=y^{\prime } \mu \left (x \right )+y \mu ^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Isolate}\hspace {3pt} \mu ^{\prime }\left (x \right ) \\ {} & {} & \mu ^{\prime }\left (x \right )=\mu \left (x \right ) \\ {} & \circ & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (x \right )={\mathrm e}^{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}\left (y \mu \left (x \right )\right )\right )d x =\int -\mu \left (x \right ) x d x +c_{1} \\ {} & \circ & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (x \right )=\int -\mu \left (x \right ) x d x +c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int -\mu \left (x \right ) x d x +c_{1}}{\mu \left (x \right )} \\ {} & \circ & \textrm {Substitute}\hspace {3pt} \mu \left (x \right )={\mathrm e}^{x} \\ {} & {} & y=\frac {\int -x \,{\mathrm e}^{x}d x +c_{1}}{{\mathrm e}^{x}} \\ {} & \circ & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {-\left (-1+x \right ) {\mathrm e}^{x}+c_{1}}{{\mathrm e}^{x}} \\ {} & \circ & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=-x +1+c_{1} {\mathrm e}^{-x} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y=\frac {x^{2}}{2}+c_{1} , y=-x +1+c_{1} {\mathrm e}^{-x}\right \} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature <- quadrature successful Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 25
dsolve(diff(y(x),x)^2+y(x)*diff(y(x),x) = x*(x+y(x)),y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {x^{2}}{2}+c_{1} \\ y \left (x \right ) &= 1+{\mathrm e}^{-x} c_{1} -x \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.057 (sec). Leaf size: 32
DSolve[(y'[x])^2+y[x]*y'[x]==x*(x+y[x]),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {x^2}{2}+c_1 \\ y(x)\to -x+c_1 e^{-x}+1 \\ \end{align*}