Internal problem ID [5321]
Internal file name [OUTPUT/4812_Friday_February_02_2024_05_13_57_AM_51034746/index.tex]
Book: Schaums Outline. Theory and problems of Differential Equations, 1st edition. Frank Ayres.
McGraw Hill 1952
Section: Chapter 6. Equations of first order and first degree (Linear equations). Supplemetary
problems. Page 39
Problem number: 23 (d).
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "abelFirstKind"
Maple gives the following as the ode type
[_Abel]
\[ \boxed {y^{\prime }+x \left (x +y\right )-x^{3} \left (x +y\right )^{3}=-1} \]
This is Abel first kind ODE, it has the form \[ y^{\prime }= f_0(x)+f_1(x) y +f_2(x)y^{2}+f_3(x)y^{3} \] Comparing the above to given ODE which is \begin {align*} y^{\prime }&=y^{3} x^{3}+3 y^{2} x^{4}+\left (3 x^{5}-x \right ) y+x^{6}-x^{2}-1\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= x^{6}-x^{2}-1\\ f_1(x) &= 3 x^{5}-x\\ f_2(x) &= 3 x^{4}\\ f_3(x) &= x^{3} \end {align*}
Since \(f_2(x)=3 x^{4}\) is not zero, then the first step is to apply the following transformation to remove \(f_2\). Let \(y = u(x) - \frac {f_2}{3 f_3}\) or \begin {align*} y &= u(x) - \left ( \frac {3 x^{4}}{3 x^{3}} \right ) \\ &= u \left (x \right )-x \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = x^{3} u \left (x \right )^{3}-x u \left (x \right )\tag {2} \end {align*}
The above ODE (2) can now be solved as separable.
Writing the ode as \begin {align*} u^{\prime }\left (x \right )&=x^{3} u^{3}-x u\\ u^{\prime }\left (x \right )&= \omega \left ( x,u\right ) \end {align*}
The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{x}+\omega \left ( \eta _{u}-\xi _{x}\right ) -\omega ^{2}\xi _{u}-\omega _{x}\xi -\omega _{u}\eta =0\tag {A} \end {align*}
The type of this ode is known. It is of type Bernoulli. Therefore we do not need
to solve the PDE (A), and can just use the lookup table shown below to find \(\xi ,\eta \)
|
ODE class |
Form |
\(\xi \) |
\(\eta \) |
|
linear ode |
\(y'=f(x) y(x) +g(x)\) |
\(0\) |
\(e^{\int fdx}\) |
|
separable ode |
\(y^{\prime }=f\left ( x\right ) g\left ( y\right ) \) |
\(\frac {1}{f}\) |
\(0\) |
|
quadrature ode |
\(y^{\prime }=f\left ( x\right ) \) |
\(0\) |
\(1\) |
|
quadrature ode |
\(y^{\prime }=g\left ( y\right ) \) |
\(1\) |
\(0\) |
|
homogeneous ODEs of Class A |
\(y^{\prime }=f\left ( \frac {y}{x}\right ) \) |
\(x\) |
\(y\) |
|
homogeneous ODEs of Class C |
\(y^{\prime }=\left ( a+bx+cy\right ) ^{\frac {n}{m}}\) |
\(1\) |
\(-\frac {b}{c}\) |
|
homogeneous class D |
\(y^{\prime }=\frac {y}{x}+g\left ( x\right ) F\left (\frac {y}{x}\right ) \) |
\(x^{2}\) |
\(xy\) |
|
First order special form ID 1 |
\(y^{\prime }=g\left ( x\right ) e^{h\left (x\right ) +by}+f\left ( x\right ) \) |
\(\frac {e^{-\int bf\left ( x\right )dx-h\left ( x\right ) }}{g\left ( x\right ) }\) |
\(\frac {f\left ( x\right )e^{-\int bf\left ( x\right ) dx-h\left ( x\right ) }}{g\left ( x\right ) }\) |
|
polynomial type ode |
\(y^{\prime }=\frac {a_{1}x+b_{1}y+c_{1}}{a_{2}x+b_{2}y+c_{2}}\) |
\(\frac {a_{1}b_{2}x-a_{2}b_{1}x-b_{1}c_{2}+b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}\) |
\(\frac {a_{1}b_{2}y-a_{2}b_{1}y-a_{1}c_{2}-a_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}\) |
|
Bernoulli ode |
\(y^{\prime }=f\left ( x\right ) y+g\left ( x\right ) y^{n}\) |
\(0\) |
\(e^{-\int \left ( n-1\right ) f\left ( x\right ) dx}y^{n}\) |
|
Reduced Riccati |
\(y^{\prime }=f_{1}\left ( x\right ) y+f_{2}\left ( x\right )y^{2}\) |
\(0\) |
\(e^{-\int f_{1}dx}\) |
|
|
|||
|
|
|||
The above table shows that \begin {align*} \xi \left (x,u\right ) &=0\\ \tag {A1} \eta \left (x,u\right ) &=u^{3} {\mathrm e}^{x^{2}} \end {align*}
The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,u\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is \begin {align*} \frac {d x}{\xi } &= \frac {d u}{\eta } = dS \tag {1} \end {align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial u}\right ) S(x,u) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Since \(\xi =0\) then in this special case \begin {align*} R = x \end {align*}
\(S\) is found from \begin {align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{u^{3} {\mathrm e}^{x^{2}}}} dy \end {align*}
Which results in \begin {align*} S&= -\frac {{\mathrm e}^{-x^{2}}}{2 u^{2}} \end {align*}
Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by evaluating \begin {align*} \frac {dS}{dR} &= \frac { S_{x} + \omega (x,u) S_{u} }{ R_{x} + \omega (x,u) R_{u} }\tag {2} \end {align*}
Where in the above \(R_{x},R_{u},S_{x},S_{u}\) are all partial derivatives and \(\omega (x,u)\) is the right hand side of the original ode given by \begin {align*} \omega (x,u) &= x^{3} u^{3}-x u \end {align*}
Evaluating all the partial derivatives gives \begin {align*} R_{x} &= 1\\ R_{u} &= 0\\ S_{x} &= \frac {{\mathrm e}^{-x^{2}} x}{u^{2}}\\ S_{u} &= \frac {{\mathrm e}^{-x^{2}}}{u^{3}} \end {align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates. \begin {align*} \frac {dS}{dR} &= {\mathrm e}^{-x^{2}} x^{3}\tag {2A} \end {align*}
We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,u\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives \begin {align*} \frac {dS}{dR} &= {\mathrm e}^{-R^{2}} R^{3} \end {align*}
The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates \(R,S\). Integrating the above gives \begin {align*} S \left (R \right ) = -\frac {\left (R^{2}+1\right ) {\mathrm e}^{-R^{2}}}{2}+c_{1}\tag {4} \end {align*}
To complete the solution, we just need to transform (4) back to \(x,u\) coordinates. This results in \begin {align*} -\frac {{\mathrm e}^{-x^{2}}}{2 u \left (x \right )^{2}} = -\frac {\left (x^{2}+1\right ) {\mathrm e}^{-x^{2}}}{2}+c_{1} \end {align*}
Which simplifies to \begin {align*} -\frac {{\mathrm e}^{-x^{2}}}{2 u \left (x \right )^{2}} = -\frac {\left (x^{2}+1\right ) {\mathrm e}^{-x^{2}}}{2}+c_{1} \end {align*}
Substituting \(u=y-x\) in the above solution gives \begin {align*} -\frac {{\mathrm e}^{-x^{2}}}{2 \left (y-x \right )^{2}} = -\frac {\left (x^{2}+1\right ) {\mathrm e}^{-x^{2}}}{2}+c_{1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} -\frac {{\mathrm e}^{-x^{2}}}{2 \left (y-x \right )^{2}} &= -\frac {\left (x^{2}+1\right ) {\mathrm e}^{-x^{2}}}{2}+c_{1} \\ \end{align*}
Verification of solutions
\[ -\frac {{\mathrm e}^{-x^{2}}}{2 \left (y-x \right )^{2}} = -\frac {\left (x^{2}+1\right ) {\mathrm e}^{-x^{2}}}{2}+c_{1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }+x \left (x +y\right )-x^{3} \left (x +y\right )^{3}=-1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-x \left (x +y\right )+x^{3} \left (x +y\right )^{3}-1 \end {array} \]
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 trying Abel <- Abel successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 109
dsolve(diff(y(x),x)+x*(x+y(x))=x^3*(x+y(x))^3-1,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -\frac {x \sqrt {{\mathrm e}^{-x^{2}} x^{2}+{\mathrm e}^{-x^{2}}+c_{1}}+{\mathrm e}^{-\frac {x^{2}}{2}}}{\sqrt {{\mathrm e}^{-x^{2}} x^{2}+{\mathrm e}^{-x^{2}}+c_{1}}} \\ y \left (x \right ) &= \frac {-x \sqrt {{\mathrm e}^{-x^{2}} x^{2}+{\mathrm e}^{-x^{2}}+c_{1}}+{\mathrm e}^{-\frac {x^{2}}{2}}}{\sqrt {{\mathrm e}^{-x^{2}} x^{2}+{\mathrm e}^{-x^{2}}+c_{1}}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 10.062 (sec). Leaf size: 85
DSolve[y'[x]+x*(x+y[x])==x^3*(x+y[x])^3-1,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -x-\frac {e^{-\frac {x^2}{2}}}{\sqrt {e^{-x^2} \left (x^2+1\right )+c_1}} \\ y(x)\to -x+\frac {e^{-\frac {x^2}{2}}}{\sqrt {e^{-x^2} \left (x^2+1\right )+c_1}} \\ y(x)\to -x \\ \end{align*}