Internal problem ID [5771]
Internal file name [OUTPUT/5019_Sunday_June_05_2022_03_17_34_PM_26513288/index.tex]
Book: Ordinary differential equations and calculus of variations. Makarets and Reshetnyak. Wold
Scientific. Singapore. 1995
Section: Chapter 1. First order differential equations. Section 1.2 Homogeneous equations
problems. page 12
Problem number: 23.
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]
\[ \boxed {y^{\prime } \left (y^{\prime }+y\right )-x \left (y+x \right )=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0] \end {align*}
The ode \begin {align*} y^{\prime } \left (y^{\prime }+y\right )-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)
This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=0\\ q(x) &=x \end {align*}
Hence the ode is \begin {align*} y^{\prime } = x \end {align*}
The domain of \(p(x)=0\) is \[
\{-\infty Integrating both sides gives \begin {align*} y &= \int { x\,\mathop {\mathrm {d}x}}\\ &= \frac {x^{2}}{2}+c_{1} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(y=0\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} 0 = c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = 0 \end {align*}
Trying the constant \begin {align*} c_{1} = 0 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=\frac {x^{2}}{2} \end {align*}
The constant \(c_{1} = 0\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {x^{2}}{2} \\
\end{align*} Verification of solutions
\[
y = \frac {x^{2}}{2}
\] Verified OK. Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {x^{2}}{2} \\
\end{align*} Verification of solutions
\[
y = \frac {x^{2}}{2}
\] Verified OK. Solving ODE (2)
This is a linear ODE. In canonical form it is written as \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 domain of \(p(x)=1\) is \[
\{-\infty Entering Linear first order ODE solver. 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 ({\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*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=0\) and \(y=0\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} 0 = 1+c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = -1 \end {align*}
Trying the constant \begin {align*} c_{2} = -1 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=1-{\mathrm e}^{-x}-x \end {align*}
The constant \(c_{2} = -1\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= 1-{\mathrm e}^{-x}-x \\
\end{align*} Verification of solutions
\[
y = 1-{\mathrm e}^{-x}-x
\] Verified OK. Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= 1-{\mathrm e}^{-x}-x \\
\end{align*} Verification of solutions
\[
y = 1-{\mathrm e}^{-x}-x
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime } \left (y^{\prime }+y\right )-x \left (y+x \right )=0, y \left (0\right )=0\right ] \\ \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} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =0 \\ {} & \circ & \textrm {Substitute}\hspace {3pt} c_{1} =0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {x^{2}}{2} \\ {} & \circ & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {x^{2}}{2} \\ \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=-x +1+c_{1} {\mathrm e}^{-x} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=1+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =-1 \\ {} & \circ & \textrm {Substitute}\hspace {3pt} c_{1} =-1\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=1-{\mathrm e}^{-x}-x \\ {} & \circ & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=1-{\mathrm e}^{-x}-x \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y=\frac {x^{2}}{2}, y=1-{\mathrm e}^{-x}-x \right \} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 9
\[
y \left (x \right ) = \frac {x^{2}}{2}
\]
✓ Solution by Mathematica
Time used: 0.043 (sec). Leaf size: 28
\begin{align*}
y(x)\to \frac {x^2}{2} \\
y(x)\to -x-e^{-x}+1 \\
\end{align*}
2.23.2 Existence and uniqueness analysis
2.23.3 Maple step by step solution
`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`
dsolve([diff(y(x),x)*(diff(y(x),x)+y(x))=x*(x+y(x)),y(0) = 0],y(x), singsol=all)
DSolve[{y'[x]*(y'[x]+y[x])==x*(x+y[x]),{y[0]==0}},y[x],x,IncludeSingularSolutions -> True]