Internal problem ID [3166]
Internal file name [OUTPUT/2658_Sunday_June_05_2022_08_38_14_AM_48662186/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: 21.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "exact"
Maple gives the following as the ode type
[_exact, _rational, [_Abel, `2nd type`, `class B`]]
\[ \boxed {-x y^{2}-2 y-\left (x^{2} y+2 x \right ) y^{\prime }=-2 x^{2}-3} \]
Entering Exact first order ODE solver. (Form one type)
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 (-y \,x^{2}-2 x\right )\mathop {\mathrm {d}y} &= \left (y^{2} x -2 x^{2}+2 y -3\right )\mathop {\mathrm {d}x}\\ \left (-y^{2} x +2 x^{2}-2 y +3\right )\mathop {\mathrm {d}x} + \left (-y \,x^{2}-2 x\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end {align*}
Comparing (1A) and (2A) shows that \begin {align*} M(x,y) &= -y^{2} x +2 x^{2}-2 y +3\\ N(x,y) &= -y \,x^{2}-2 x \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^{2} x +2 x^{2}-2 y +3\right )\\ &= -2 x y -2 \end {align*}
And \begin {align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (-y \,x^{2}-2 x\right )\\ &= -2 x y -2 \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 -y^{2} x +2 x^{2}-2 y +3\mathop {\mathrm {d}x} \\ \tag{3} \phi &= \frac {2 x^{3}}{3}-\frac {x^{2} y^{2}}{2}+\left (-2 y +3\right ) 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{equation} \tag{4} \frac {\partial \phi }{\partial y} = -y \,x^{2}-2 x+f'(y) \end{equation} But equation (2) says that \(\frac {\partial \phi }{\partial y} = -y \,x^{2}-2 x\). Therefore equation (4) becomes \begin{equation} \tag{5} -y \,x^{2}-2 x = -y \,x^{2}-2 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 \(\phi \) \[ \phi = \frac {2 x^{3}}{3}-\frac {x^{2} y^{2}}{2}+\left (-2 y +3\right ) x+ c_{1} \] 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 \[ c_{1} = \frac {2 x^{3}}{3}-\frac {x^{2} y^{2}}{2}+\left (-2 y +3\right ) x \]
The solution(s) found are the following \begin{align*} \tag{1} \frac {2 x^{3}}{3}-\frac {y^{2} x^{2}}{2}+\left (-2 y+3\right ) x &= c_{1} \\ \end{align*}
Verification of solutions
\[ \frac {2 x^{3}}{3}-\frac {y^{2} x^{2}}{2}+\left (-2 y+3\right ) x = c_{1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -x y^{2}-2 y-\left (x^{2} y+2 x \right ) y^{\prime }=-2 x^{2}-3 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \square & {} & \textrm {Check if ODE is exact}\hspace {3pt} \\ {} & \circ & \textrm {ODE is exact if the lhs is the total derivative of a}\hspace {3pt} C^{2}\hspace {3pt}\textrm {function}\hspace {3pt} \\ {} & {} & F^{\prime }\left (x , y\right )=0 \\ {} & \circ & \textrm {Compute derivative of lhs}\hspace {3pt} \\ {} & {} & F^{\prime }\left (x , y\right )+\left (\frac {\partial }{\partial y}F \left (x , y\right )\right ) y^{\prime }=0 \\ {} & \circ & \textrm {Evaluate derivatives}\hspace {3pt} \\ {} & {} & -2 x y -2=-2 x y -2 \\ {} & \circ & \textrm {Condition met, ODE is exact}\hspace {3pt} \\ \bullet & {} & \textrm {Exact ODE implies solution will be of this form}\hspace {3pt} \\ {} & {} & \left [F \left (x , y\right )=c_{1} , M \left (x , y\right )=F^{\prime }\left (x , y\right ), N \left (x , y\right )=\frac {\partial }{\partial y}F \left (x , y\right )\right ] \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} F \left (x , y\right )\hspace {3pt}\textrm {by integrating}\hspace {3pt} M \left (x , y\right )\hspace {3pt}\textrm {with respect to}\hspace {3pt} x \\ {} & {} & F \left (x , y\right )=\int \left (-y^{2} x +2 x^{2}-2 y +3\right )d x +f_{1} \left (y \right ) \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & F \left (x , y\right )=-\frac {x^{2} y^{2}}{2}+\frac {2 x^{3}}{3}-2 x y +3 x +f_{1} \left (y \right ) \\ \bullet & {} & \textrm {Take derivative of}\hspace {3pt} F \left (x , y\right )\hspace {3pt}\textrm {with respect to}\hspace {3pt} y \\ {} & {} & N \left (x , y\right )=\frac {\partial }{\partial y}F \left (x , y\right ) \\ \bullet & {} & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & -y \,x^{2}-2 x =-y \,x^{2}-2 x +\frac {d}{d y}f_{1} \left (y \right ) \\ \bullet & {} & \textrm {Isolate for}\hspace {3pt} \frac {d}{d y}f_{1} \left (y \right ) \\ {} & {} & \frac {d}{d y}f_{1} \left (y \right )=0 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} f_{1} \left (y \right ) \\ {} & {} & f_{1} \left (y \right )=0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} f_{1} \left (y \right )\hspace {3pt}\textrm {into equation for}\hspace {3pt} F \left (x , y\right ) \\ {} & {} & F \left (x , y\right )=-\frac {1}{2} x^{2} y^{2}+\frac {2}{3} x^{3}-2 x y +3 x \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} F \left (x , y\right )\hspace {3pt}\textrm {into the solution of the ODE}\hspace {3pt} \\ {} & {} & -\frac {1}{2} x^{2} y^{2}+\frac {2}{3} x^{3}-2 x y +3 x =c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\frac {-2-\frac {\sqrt {12 x^{3}-18 c_{1} +54 x +36}}{3}}{x}, y=\frac {-2+\frac {\sqrt {12 x^{3}-18 c_{1} +54 x +36}}{3}}{x}\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 Chini differential order: 1; looking for linear symmetries trying exact <- exact successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 53
dsolve((2*x^2-x*y(x)^2-2*y(x)+3)-(x^2*y(x)+2*x)*diff(y(x),x)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {-6-\sqrt {12 x^{3}+18 c_{1} +54 x +36}}{3 x} \\ y \left (x \right ) &= \frac {-6+\sqrt {12 x^{3}+18 c_{1} +54 x +36}}{3 x} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.646 (sec). Leaf size: 87
DSolve[(2*x^2-x*y[x]^2-2*y[x]+3)-(x^2*y[x]+2*x)*y'[x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {6 x+\sqrt {3} \sqrt {x^2 \left (4 x^3+18 x+12+3 c_1\right )}}{3 x^2} \\ y(x)\to \frac {-6 x+\sqrt {3} \sqrt {x^2 \left (4 x^3+18 x+12+3 c_1\right )}}{3 x^2} \\ \end{align*}