problem 29

Internal problem ID [3174]
Internal ﬁle name [OUTPUT/2666_Sunday_June_05_2022_08_38_33_AM_32086448/index.tex]

Book: Diﬀerential equations for engineers by Wei-Chau XIE, Cambridge Press 2010
Section: Chapter 2. First-Order and Simple Higher-Order Diﬀerential Equations. Page 78
Problem number: 29.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "homogeneousTypeD2", "exactWithIntegrationFactor"

1.29.1 Solving as homogeneousTypeD2 ode

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*} u \left (x \right ) x \left (2 x -u \left (x \right ) x +2\right )+2 \left (x -u \left (x \right ) x \right ) \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) = 0 \end {align*}

In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {\left (x +2\right ) u \left (u -2\right )}{x \left (2 u -2\right )} \end {align*}

Where \(f(x)=-\frac {x +2}{x}\) and \(g(u)=\frac {u \left (u -2\right )}{2 u -2}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u \left (u -2\right )}{2 u -2}} \,du &= -\frac {x +2}{x} \,d x \\ \int { \frac {1}{\frac {u \left (u -2\right )}{2 u -2}} \,du} &= \int {-\frac {x +2}{x} \,d x} \\ \ln \left (u \left (u -2\right )\right )&=-x -2 \ln \left (x \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} u \left (u -2\right ) &= {\mathrm e}^{-x -2 \ln \left (x \right )+c_{2}} \end {align*}

Which simpliﬁes to \begin {align*} u \left (u -2\right ) &= c_{3} {\mathrm e}^{-x -2 \ln \left (x \right )} \end {align*}

Which simpliﬁes to \[ u \left (x \right ) \left (u \left (x \right )-2\right ) = \frac {c_{3} {\mathrm e}^{-x} {\mathrm e}^{c_{2}}}{x^{2}} \] The solution is \[ u \left (x \right ) \left (u \left (x \right )-2\right ) = \frac {c_{3} {\mathrm e}^{-x} {\mathrm e}^{c_{2}}}{x^{2}} \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \frac {y \left (\frac {y}{x}-2\right )}{x} = \frac {c_{3} {\mathrm e}^{-x} {\mathrm e}^{c_{2}}}{x^{2}}\\ \frac {-2 y x +y^{2}}{x^{2}} = \frac {c_{3} {\mathrm e}^{c_{2} -x}}{x^{2}} \end {align*}

Which simpliﬁes to \begin {align*} -2 y x +y^{2} = c_{3} {\mathrm e}^{c_{2} -x} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} -2 y x +y^{2} &= c_{3} {\mathrm e}^{c_{2} -x} \\ \end{align*}

Veriﬁcation of solutions

\[ -2 y x +y^{2} = c_{3} {\mathrm e}^{c_{2} -x} \] Veriﬁed OK.

1.29.2 Solving as exact ode

To solve an ode of the form\begin {equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A} \end {equation} We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisﬁes the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \] Hence\begin {equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B} \end {equation} Comparing (A,B) shows that\begin {align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end {align*}

But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\] If the above condition is satisﬁed, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisﬁed. If this condition is not satisﬁed then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The ﬁrst step is to write the ODE in standard form to check for exactness, which is \[ M(x,y) \mathop {\mathrm {d}x}+ N(x,y) \mathop {\mathrm {d}y}=0 \tag {1A} \] Therefore \begin {align*} \left (2 x -2 y\right )\mathop {\mathrm {d}y} &= \left (-y \left (2 x -y +2\right )\right )\mathop {\mathrm {d}x}\\ \left (y \left (2 x -y +2\right )\right )\mathop {\mathrm {d}x} + \left (2 x -2 y\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end {align*}

Comparing (1A) and (2A) shows that \begin {align*} M(x,y) &= y \left (2 x -y +2\right )\\ N(x,y) &= 2 x -2 y \end {align*}

The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisﬁed \[ \frac {\partial M}{\partial y} = \frac {\partial N}{\partial x} \] Using result found above gives \begin {align*} \frac {\partial M}{\partial y} &= \frac {\partial }{\partial y} \left (y \left (2 x -y +2\right )\right )\\ &= 2 x -2 y +2 \end {align*}

And \begin {align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (2 x -2 y\right )\\ &= 2 \end {align*}

Since \(\frac {\partial M}{\partial y} \neq \frac {\partial N}{\partial x}\), then the ODE is not exact. Since the ODE is not exact, we will try to ﬁnd an integrating factor to make it exact. Let \begin {align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial y} - \frac {\partial N}{\partial x} \right ) \\ &=\frac {1}{2 x -2 y}\left ( \left ( 2 x -2 y +2\right ) - \left (2 \right ) \right ) \\ &=1 \end {align*}

Since \(A\) does not depend on \(y\), then it can be used to ﬁnd an integrating factor. The integrating factor \(\mu \) is \begin {align*} \mu &= e^{ \int A \mathop {\mathrm {d}x} } \\ &= e^{\int 1\mathop {\mathrm {d}x} } \end {align*}

The result of integrating gives \begin {align*} \mu &= e^{x } \\ &= {\mathrm e}^{x} \end {align*}

\(M\) and \(N\) are multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\) and \(\overline {N}\) for now so not to confuse them with the original \(M\) and \(N\). \begin {align*} \overline {M} &=\mu M \\ &= {\mathrm e}^{x}\left (y \left (2 x -y +2\right )\right ) \\ &= y \left (2 x -y +2\right ) {\mathrm e}^{x} \end {align*}

And \begin {align*} \overline {N} &=\mu N \\ &= {\mathrm e}^{x}\left (2 x -2 y\right ) \\ &= 2 \left (x -y \right ) {\mathrm e}^{x} \end {align*}

Now a modiﬁed ODE is ontained from the original ODE, which is exact and can be solved. The modiﬁed ODE is \begin {align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \\ \left (y \left (2 x -y +2\right ) {\mathrm e}^{x}\right ) + \left (2 \left (x -y \right ) {\mathrm e}^{x}\right ) \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \end {align*}

The following equations are now set up to solve for the function \(\phi \left (x,y\right )\) \begin {align*} \frac {\partial \phi }{\partial x } &= \overline {M}\tag {1} \\ \frac {\partial \phi }{\partial y } &= \overline {N}\tag {2} \end {align*}

Integrating (1) w.r.t. \(x\) gives \begin{align*} \int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int \overline {M}\mathop {\mathrm {d}x} \\ \int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int y \left (2 x -y +2\right ) {\mathrm e}^{x}\mathop {\mathrm {d}x} \\ \tag{3} \phi &= \left (2 x -y \right ) y \,{\mathrm e}^{x}+ f(y) \\ \end{align*} Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function of both \(x\) and \(y\). Taking derivative of equation (3) w.r.t \(y\) gives \begin{align*} \tag{4} \frac {\partial \phi }{\partial y} &= -y \,{\mathrm e}^{x}+\left (2 x -y \right ) {\mathrm e}^{x}+f'(y) \\ &=2 \left (x -y \right ) {\mathrm e}^{x}+f'(y) \\ \end{align*} But equation (2) says that \(\frac {\partial \phi }{\partial y} = 2 \left (x -y \right ) {\mathrm e}^{x}\). Therefore equation (4) becomes \begin{equation} \tag{5} 2 \left (x -y \right ) {\mathrm e}^{x} = 2 \left (x -y \right ) {\mathrm e}^{x}+f'(y) \end{equation} Solving equation (5) for \( f'(y)\) gives \[ f'(y) = 0 \] Therefore \[ f(y) = c_{1} \] Where \(c_{1}\) is constant of integration. Substituting this result for \(f(y)\) into equation (3) gives \(\phi \) \[ \phi = \left (2 x -y \right ) y \,{\mathrm e}^{x}+ c_{1} \] But since \(\phi \) itself is a constant function, then let \(\phi =c_{2}\) where \(c_{2}\) is new constant and combining \(c_{1}\) and \(c_{2}\) constants into new constant \(c_{1}\) gives the solution as \[ c_{1} = \left (2 x -y \right ) y \,{\mathrm e}^{x} \]

Summary

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

Veriﬁcation of solutions

\[ \left (2 x -y\right ) y \,{\mathrm e}^{x} = c_{1} \] Veriﬁed OK.

1.29.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y \left (2 x -y+2\right )+2 \left (x -y\right ) y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {y \left (2 x -y+2\right )}{2 \left (x -y\right )} \end {array} \]

Maple trace

`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`

\begin{align*} y \left (x \right ) &= \frac {c_{1} x -\sqrt {{\mathrm e}^{x} c_{1} \left ({\mathrm e}^{x} c_{1} x^{2}+1\right )}\, {\mathrm e}^{-x}}{c_{1}} \\ y \left (x \right ) &= \frac {c_{1} x +\sqrt {{\mathrm e}^{x} c_{1} \left ({\mathrm e}^{x} c_{1} x^{2}+1\right )}\, {\mathrm e}^{-x}}{c_{1}} \\ \end{align*}

\begin{align*} y(x)\to x-e^{-x} \sqrt {e^x \left (e^x x^2-e^{2 c_1}\right )} \\ y(x)\to x+e^{-x} \sqrt {e^x \left (e^x x^2-e^{2 c_1}\right )} \\ y(x)\to x-e^{-x} \sqrt {e^{2 x} x^2} \\ y(x)\to e^{-x} \sqrt {e^{2 x} x^2}+x \\ \end{align*}

1.29 problem 29

1.29.1 Solving as homogeneousTypeD2 ode

1.29.2 Solving as exact ode

1.29.3 Maple step by step solution