Internal problem ID [2322]
Internal file name [OUTPUT/2322_Tuesday_February_27_2024_08_26_05_AM_72160901/index.tex]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 37, page 171
Problem number: 9.
ODE order: 1.
ODE degree: 3.
The type(s) of ODE detected by this program : "exact", "linear", "quadrature", "separable", "homogeneousTypeD2", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[_quadrature]
\[ \boxed {{y^{\prime }}^{3}+\left (x +y-2 y x \right ) {y^{\prime }}^{2}-2 y^{\prime } x y \left (x +y\right )=0} \] The ode \begin {align*} {y^{\prime }}^{3}+\left (x +y-2 y x \right ) {y^{\prime }}^{2}-2 y^{\prime } x y \left (x +y\right ) = 0 \end {align*}
is factored to \begin {align*} y^{\prime } \left (x +y+y^{\prime }\right ) \left (2 y x -y^{\prime }\right ) = 0 \end {align*}
Which gives the following equations \begin {align*} y^{\prime } = 0\tag {1} \\ x +y+y^{\prime } = 0\tag {2} \\ 2 y x -y^{\prime } = 0\tag {3} \\ \end {align*}
Each of the above equations is now solved.
Solving ODE (1) Integrating both sides gives \begin {align*} y &= \int { 0\,\mathop {\mathrm {d}x}}\\ &= c_{1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \] 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 ({\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.
Solving ODE (3) In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= 2 y x \end {align*}
Where \(f(x)=2 x\) and \(g(y)=y\). Integrating both sides gives \begin {align*} \frac {1}{y} \,dy &= 2 x \,d x\\ \int { \frac {1}{y} \,dy} &= \int {2 x \,d x}\\ \ln \left (y \right )&=x^{2}+c_{3}\\ y&={\mathrm e}^{x^{2}+c_{3}}\\ &=c_{3} {\mathrm e}^{x^{2}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{3} {\mathrm e}^{x^{2}} \\ \end{align*}
Verification of solutions
\[ y = c_{3} {\mathrm e}^{x^{2}} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{3} {\mathrm e}^{x^{2}} \\ \end{align*}
Verification of solutions
\[ y = c_{3} {\mathrm e}^{x^{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & {y^{\prime }}^{3}+\left (x +y-2 y x \right ) {y^{\prime }}^{2}-2 y^{\prime } x y \left (x +y\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 }=0, y^{\prime }=-x -y, y^{\prime }=2 y x \right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=0 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int 0d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=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=1-x +c_{1} {\mathrm e}^{-x} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=2 y x \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=2 x \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int 2 x d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=x^{2}+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{x^{2}+c_{1}} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y=c_{1} , y=1-x +c_{1} {\mathrm e}^{-x}, y={\mathrm e}^{x^{2}+c_{1}}\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 <- 1st order linear 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: 28
dsolve(diff(y(x),x)^3+(x+y(x)-2*x*y(x))*diff(y(x),x)^2-2*diff(y(x),x)*x*y(x)*(x+y(x))=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= {\mathrm e}^{x^{2}} c_{1} \\ y \left (x \right ) &= 1+{\mathrm e}^{-x} c_{1} -x \\ y \left (x \right ) &= c_{1} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.06 (sec). Leaf size: 36
DSolve[y'[x]^3+(x+y[x]-2*x*y[x])*y'[x]^2-2*y'[x]*x*y[x]*(x+y[x])==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to c_1 \\ y(x)\to c_1 e^{x^2} \\ y(x)\to -x+c_1 e^{-x}+1 \\ \end{align*}