Internal problem ID [15090]
Internal file name [OUTPUT/15091_Sunday_April_21_2024_01_29_34_PM_85331927/index.tex]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Section 8. First order not solved for the derivative. Exercises page 67
Problem number: 203.
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]
\[ \boxed {{y^{\prime }}^{2}-\left (y+2 x \right ) y^{\prime }+y x=-x^{2}} \] The ode \begin {align*} {y^{\prime }}^{2}-\left (y+2 x \right ) y^{\prime }+y x = -x^{2} \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 \left (-1\right )d 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 ({\mathrm e}^{-x} y\right ) &= \left ({\mathrm e}^{-x}\right ) \left (x\right )\\ \mathrm {d} \left ({\mathrm e}^{-x} y\right ) &= \left (x \,{\mathrm e}^{-x}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} {\mathrm e}^{-x} y &= \int {x \,{\mathrm e}^{-x}\,\mathrm {d} x}\\ {\mathrm e}^{-x} y &= -\left (x +1\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 (x +1\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}-\left (y+2 x \right ) y^{\prime }+y x =-x^{2} \\ \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 (x +1\right ) {\mathrm e}^{-x}+c_{1}}{{\mathrm e}^{-x}} \\ {} & \circ & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{x}-x -1 \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y=\frac {x^{2}}{2}+c_{1} , y=c_{1} {\mathrm e}^{x}-x -1\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: 23
dsolve(diff(y(x),x)^2-(2*x+y(x))*diff(y(x),x)+x^2+x*y(x)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {x^{2}}{2}+c_{1} \\ y \left (x \right ) &= -x -1+{\mathrm e}^{x} c_{1} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.058 (sec). Leaf size: 30
DSolve[y'[x]^2-(2*x+y[x])*y'[x]+x^2+x*y[x]==0,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*}