6.19 problem 19

Internal problem ID [2005]
Internal file name [OUTPUT/2005_Sunday_February_25_2024_06_44_29_AM_23231500/index.tex]

Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath. Boston. 1964
Section: Exercise 10, page 41
Problem number: 19.
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, [_1st_order, `_with_symmetry_[F(x)*G(y),0]`]]

\[ \boxed {y^{2}+\left (2 x y-y^{2}\right ) y^{\prime }=-1} \] With initial conditions \begin {align*} [y \left (0\right ) = -1] \end {align*}

6.19.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= \frac {y^{2}+1}{y \left (-2 x +y \right )} \end {align*}

The \(x\) domain of \(f(x,y)\) when \(y=-1\) is \[ \left \{x <-\frac {1}{2}\boldsymbol {\lor }-\frac {1}{2}

The \(x\) domain of \(\frac {\partial f}{\partial y}\) when \(y=-1\) is \[ \left \{x <-\frac {1}{2}\boldsymbol {\lor }-\frac {1}{2}

6.19.2 Solving as exact ode

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

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

And \begin {align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (2 x y -y^{2}\right )\\ &= 2 y \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}+1\mathop {\mathrm {d}x} \\ \tag{3} \phi &= \left (y^{2}+1\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} = 2 x y+f'(y) \end{equation} But equation (2) says that \(\frac {\partial \phi }{\partial y} = 2 x y -y^{2}\). Therefore equation (4) becomes \begin{equation} \tag{5} 2 x y -y^{2} = 2 x y+f'(y) \end{equation} Solving equation (5) for \( f'(y)\) gives \[ f'(y) = -y^{2} \] Integrating the above w.r.t \(y\) gives \begin{align*} \int f'(y) \mathop {\mathrm {d}y} &= \int \left ( -y^{2}\right ) \mathop {\mathrm {d}y} \\ f(y) &= -\frac {y^{3}}{3}+ c_{1} \\ \end{align*} Where \(c_{1}\) is constant of integration. Substituting result found above for \(f(y)\) into equation (3) gives \(\phi \) \[ \phi = \left (y^{2}+1\right ) x -\frac {y^{3}}{3}+ c_{1} \] 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 \[ c_{1} = \left (y^{2}+1\right ) x -\frac {y^{3}}{3} \] Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(y=-1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} {\frac {1}{3}} = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = {\frac {1}{3}} \end {align*}

Trying the constant \begin {align*} c_{1} = {\frac {1}{3}} \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \left (y^{2}+1\right ) x -\frac {y^{3}}{3} = {\frac {1}{3}} \end {align*}

The constant \(c_{1} = {\frac {1}{3}}\) gives valid solution.

Summary

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

Verification of solutions

\[ y^{2} x +x -\frac {y^{3}}{3} = {\frac {1}{3}} \] Verified OK.

Maple trace

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

✓ Solution by Maple

Time used: 0.313 (sec). Leaf size: 137

dsolve([(y(x)^2+1)+(2*x*y(x)-y(x)^2)*diff(y(x),x)=0,y(0) = -1],y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {\left (-4+12 x +8 x^{3}+4 \sqrt {12 x^{4}-4 x^{3}+9 x^{2}-6 x +1}\right )^{\frac {1}{3}} \left (i \sqrt {3}-1\right )}{4}-\frac {\left (i \sqrt {3}\, x +x -\left (-4+12 x +8 x^{3}+4 \sqrt {12 x^{4}-4 x^{3}+9 x^{2}-6 x +1}\right )^{\frac {1}{3}}\right ) x}{\left (-4+12 x +8 x^{3}+4 \sqrt {12 x^{4}-4 x^{3}+9 x^{2}-6 x +1}\right )^{\frac {1}{3}}} \]

✓ Solution by Mathematica

Time used: 4.803 (sec). Leaf size: 100

DSolve[{(y[x]^2+1)+(2*x*y[x]-y[x]^2)*y'[x]==0,{y[0]==-1}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to -\frac {\sqrt [3]{2} x^2}{\sqrt [3]{-2 x^3+\sqrt {12 x^4-4 x^3+9 x^2-6 x+1}-3 x+1}}-\frac {\sqrt [3]{-2 x^3+\sqrt {12 x^4-4 x^3+9 x^2-6 x+1}-3 x+1}}{\sqrt [3]{2}}+x \]