Internal problem ID [6778]
Internal file name [OUTPUT/6025_Monday_July_25_2022_01_59_52_AM_16739784/index.tex]
Book: Elementary differential equations. By Earl D. Rainville, Phillip E. Bedient. Macmilliam
Publishing Co. NY. 6th edition. 1981.
Section: CHAPTER 16. Nonlinear equations. Section 94. Factoring the left member. EXERCISES
Page 309
Problem number: 12.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "quadrature", "homogeneousTypeD2", "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[_quadrature]
\[ \boxed {\left (4 x -y\right ) {y^{\prime }}^{2}+6 \left (x -y\right ) y^{\prime }-5 y=-2 x} \] The ode \begin {align*} \left (4 x -y\right ) {y^{\prime }}^{2}+6 \left (x -y\right ) y^{\prime }-5 y = -2 x \end {align*}
is factored to \begin {align*} \left (y^{\prime }+1\right ) \left (y^{\prime } y-4 y^{\prime } x +5 y-2 x \right ) = 0 \end {align*}
Which gives the following equations \begin {align*} y^{\prime }+1 = 0\tag {1} \\ y^{\prime } y-4 y^{\prime } x +5 y-2 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 { -1\,\mathop {\mathrm {d}x}}\\ &= -x +c_{1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -x +c_{1} \\ \end{align*}
Verification of solutions
\[ y = -x +c_{1} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -x +c_{1} \\ \end{align*}
Verification of solutions
\[ y = -x +c_{1} \] Verified OK.
Solving ODE (2) Using the change of variables \(y = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) u \left (x \right ) x -4 \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) x +5 u \left (x \right ) x = 2 x \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {u^{2}+u -2}{x \left (u -4\right )} \end {align*}
Where \(f(x)=-\frac {1}{x}\) and \(g(u)=\frac {u^{2}+u -2}{u -4}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{2}+u -2}{u -4}} \,du &= -\frac {1}{x} \,d x \\ \int { \frac {1}{\frac {u^{2}+u -2}{u -4}} \,du} &= \int {-\frac {1}{x} \,d x} \\ 2 \ln \left (u +2\right )-\ln \left (u -1\right )&=-\ln \left (x \right )+c_{3} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\mathrm e}^{2 \ln \left (u +2\right )-\ln \left (u -1\right )} &= {\mathrm e}^{-\ln \left (x \right )+c_{3}} \end {align*}
Which simplifies to \begin {align*} \frac {\left (u +2\right )^{2}}{u -1} &= \frac {c_{4}}{x} \end {align*}
The solution is \[ \frac {\left (u \left (x \right )+2\right )^{2}}{u \left (x \right )-1} = \frac {c_{4}}{x} \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \frac {\left (\frac {y}{x}+2\right )^{2}}{\frac {y}{x}-1} = \frac {c_{4}}{x}\\ \frac {\left (y+2 x \right )^{2}}{x \left (y-x \right )} = \frac {c_{4}}{x} \end {align*}
Which simplifies to \begin {align*} -\frac {\left (y+2 x \right )^{2}}{x -y} = c_{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} -\frac {\left (y+2 x \right )^{2}}{x -y} &= c_{4} \\ \end{align*}
Verification of solutions
\[ -\frac {\left (y+2 x \right )^{2}}{x -y} = c_{4} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} -\frac {\left (y+2 x \right )^{2}}{x -y} &= c_{4} \\ \end{align*}
Verification of solutions
\[ -\frac {\left (y+2 x \right )^{2}}{x -y} = c_{4} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (4 x -y\right ) {y^{\prime }}^{2}+6 \left (x -y\right ) y^{\prime }-5 y=-2 x \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=-1, y^{\prime }=-\frac {-2 x +5 y}{-4 x +y}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int \left (-1\right )d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=-x +c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-x +c_{1} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {-2 x +5 y}{-4 x +y} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , y=-x +c_{1} \right \} \end {array} \]
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 trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying homogeneous D <- homogeneous successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 55
dsolve((4*x-y(x))*diff(y(x),x)^2+6*(x-y(x))*diff(y(x),x)+2*x-5*y(x)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -x +c_{1} \\ y \left (x \right ) &= \frac {-4 c_{1} x +\sqrt {-12 c_{1} x +1}+1}{2 c_{1}} \\ y \left (x \right ) &= \frac {-4 c_{1} x -\sqrt {-12 c_{1} x +1}+1}{2 c_{1}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 1.077 (sec). Leaf size: 90
DSolve[(4*x-y[x])*(y'[x])^2+6*(x-y[x])*y'[x]+2*x-5*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {1}{2} \left (-4 x-e^{\frac {c_1}{2}} \sqrt {12 x+e^{c_1}}-e^{c_1}\right ) \\ y(x)\to \frac {1}{2} \left (-4 x+e^{\frac {c_1}{2}} \sqrt {12 x+e^{c_1}}-e^{c_1}\right ) \\ y(x)\to -x+c_1 \\ \end{align*}