problem Problem 5

Internal problem ID [2717]
Internal ﬁle name [OUTPUT/2209_Sunday_June_05_2022_02_54_54_AM_50811869/index.tex]

Book: Diﬀerential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section: Chapter 1, First-Order Diﬀerential Equations. Section 1.9, Exact Diﬀerential Equations. page 91
Problem number: Problem 5.
ODE order: 1.
ODE degree: 1.

5.5.1 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 (-\frac {1}{2 y}\right )\mathop {\mathrm {d}y} &= \left (\frac {x}{x^{2}+1}\right )\mathop {\mathrm {d}x}\\ \left (-\frac {x}{x^{2}+1}\right )\mathop {\mathrm {d}x} + \left (-\frac {1}{2 y}\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end {align*}

Comparing (1A) and (2A) shows that \begin {align*} M(x,y) &= -\frac {x}{x^{2}+1}\\ N(x,y) &= -\frac {1}{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 (-\frac {x}{x^{2}+1}\right )\\ &= 0 \end {align*}

And \begin {align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (-\frac {1}{2 y}\right )\\ &= 0 \end {align*}

Since \(\frac {\partial M}{\partial y}= \frac {\partial N}{\partial x}\), then the ODE is exact The following equations are now set up to solve for the function \(\phi \left (x,y\right )\) \begin {align*} \frac {\partial \phi }{\partial x } &= M\tag {1} \\ \frac {\partial \phi }{\partial y } &= 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 M\mathop {\mathrm {d}x} \\ \int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int -\frac {x}{x^{2}+1}\mathop {\mathrm {d}x} \\ \tag{3} \phi &= -\frac {\ln \left (x^{2}+1\right )}{2}+ 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{equation} \tag{4} \frac {\partial \phi }{\partial y} = 0+f'(y) \end{equation} But equation (2) says that \(\frac {\partial \phi }{\partial y} = -\frac {1}{2 y}\). Therefore equation (4) becomes \begin{equation} \tag{5} -\frac {1}{2 y} = 0+f'(y) \end{equation} Solving equation (5) for \( f'(y)\) gives \[ f'(y) = -\frac {1}{2 y} \] Integrating the above w.r.t \(y\) gives \begin{align*} \int f'(y) \mathop {\mathrm {d}y} &= \int \left ( -\frac {1}{2 y}\right ) \mathop {\mathrm {d}y} \\ f(y) &= -\frac {\ln \left (y \right )}{2}+ c_{1} \\ \end{align*} Where \(c_{1}\) is constant of integration. Substituting result found above for \(f(y)\) into equation (3) gives \(\phi \) \[ \phi = -\frac {\ln \left (x^{2}+1\right )}{2}-\frac {\ln \left (y \right )}{2}+ c_{1} \] 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 \[ c_{1} = -\frac {\ln \left (x^{2}+1\right )}{2}-\frac {\ln \left (y \right )}{2} \] The solution becomes\[ y = \frac {{\mathrm e}^{-2 c_{1}}}{x^{2}+1} \]

Summary

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

Veriﬁcation of solutions

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

5.5.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 y x +\left (x^{2}+1\right ) y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=-\frac {2 x}{x^{2}+1} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int -\frac {2 x}{x^{2}+1}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=-\ln \left (x^{2}+1\right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {{\mathrm e}^{c_{1}}}{x^{2}+1} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
<- 1st order linear successful`

5.5 problem Problem 5

5.5.1 Solving as exact ode

5.5.2 Maple step by step solution