13.6 problem Ex 6

Internal problem ID [11204]
Internal file name [OUTPUT/10190_Tuesday_December_06_2022_03_59_41_AM_54585174/index.tex]

Book: An elementary treatise on differential equations by Abraham Cohen. DC heath publishers. 1906
Section: Chapter IV, differential equations of the first order and higher degree than the first. Article 24. Equations solvable for \(p\). Page 49
Problem number: Ex 6.
ODE order: 1.
ODE degree: 3.

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]

\[ \boxed {{y^{\prime }}^{3}-\left (2 x +y^{2}\right ) {y^{\prime }}^{2}+\left (x^{2}-y^{2}+2 y^{2} x \right ) y^{\prime }-\left (-y^{2}+x^{2}\right ) y^{2}=0} \] The ode \begin {align*} {y^{\prime }}^{3}-\left (2 x +y^{2}\right ) {y^{\prime }}^{2}+\left (x^{2}-y^{2}+2 y^{2} x \right ) y^{\prime }-\left (-y^{2}+x^{2}\right ) y^{2} = 0 \end {align*}

is factored to \begin {align*} \left (y^{2}-y^{\prime }\right ) \left (y^{\prime }+y-x \right ) \left (-y^{\prime }+y+x \right ) = 0 \end {align*}

Which gives the following equations \begin {align*} y^{2}-y^{\prime } = 0\tag {1} \\ y^{\prime }+y-x = 0\tag {2} \\ -y^{\prime }+y+x = 0\tag {3} \\ \end {align*}

Each of the above equations is now solved.

Solving ODE (1) Integrating both sides gives \begin {align*} \int \frac {1}{y^{2}}d y &= x +c_{1}\\ -\frac {1}{y}&=x +c_{1} \end {align*}

Solving for \(y\) gives these solutions \begin {align*} y_1&=-\frac {1}{x +c_{1}} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {1}{x +c_{1}} \\ \end{align*}

Verification of solutions

\[ y = -\frac {1}{x +c_{1}} \] Verified OK.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {1}{x +c_{1}} \\ \end{align*}

Verification of solutions

\[ y = -\frac {1}{x +c_{1}} \] 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 &= x -1+c_{2} {\mathrm e}^{-x} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= x -1+c_{2} {\mathrm e}^{-x} \\ \end{align*}

Verification of solutions

\[ y = x -1+c_{2} {\mathrm e}^{-x} \] Verified OK.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= x -1+c_{2} {\mathrm e}^{-x} \\ \end{align*}

Verification of solutions

\[ y = x -1+c_{2} {\mathrm e}^{-x} \] Verified OK.

Solving ODE (3)

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_{3} \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_{3} {\mathrm e}^{x} \end {align*}

which simplifies to \begin {align*} y &= -x -1+c_{3} {\mathrm e}^{x} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -x -1+c_{3} {\mathrm e}^{x} \\ \end{align*}

Verification of solutions

\[ y = -x -1+c_{3} {\mathrm e}^{x} \] Verified OK.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -x -1+c_{3} {\mathrm e}^{x} \\ \end{align*}

Verification of solutions

\[ y = -x -1+c_{3} {\mathrm e}^{x} \] Verified OK.

13.6.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & {y^{\prime }}^{3}-\left (2 x +y^{2}\right ) {y^{\prime }}^{2}+\left (x^{2}-y^{2}+2 y^{2} x \right ) y^{\prime }-\left (-y^{2}+x^{2}\right ) y^{2}=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 }=y^{2}, y^{\prime }=x -y, y^{\prime }=y+x \right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=y^{2} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y^{2}}=1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y^{2}}d x =\int 1d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{y}=x +c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-\frac {1}{x +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=x -1+c_{1} {\mathrm e}^{-x} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=y+x \\ {} & \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 {1}{x +c_{1}}, y=x -1+c_{1} {\mathrm e}^{-x}, y=c_{1} {\mathrm e}^{x}-x -1\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 
<- 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: 35

dsolve(diff(y(x),x)^3-(2*x+y(x)^2)*diff(y(x),x)^2+(x^2-y(x)^2+2*x*y(x)^2)*diff(y(x),x)-(x^2-y(x)^2)*y(x)^2=0,y(x), singsol=all)
 

\begin{align*} y \left (x \right ) &= \frac {1}{c_{1} -x} \\ y \left (x \right ) &= -x -1+c_{1} {\mathrm e}^{x} \\ y \left (x \right ) &= x -1+c_{1} {\mathrm e}^{-x} \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.276 (sec). Leaf size: 48

DSolve[(y'[x])^3-(2*x+y[x]^2)*(y'[x])^2+(x^2-y[x]^2+2*x*y[x]^2)*y'[x]-(x^2-y[x]^2)*y[x]^2==0,y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to -\frac {1}{x+c_1} \\ y(x)\to x+c_1 e^{-x}-1 \\ y(x)\to -x+c_1 e^x-1 \\ y(x)\to 0 \\ \end{align*}