Internal problem ID [7384]
Internal file name [OUTPUT/6485_Saturday_August_06_2022_05_19_37_AM_49721640/index.tex
]
Book: First order enumerated odes
Section: section 1
Problem number: 68.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "first order special form ID 1", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[[_1st_order, `_with_symmetry_[F(x),G(x)]`]]
\[ \boxed {y^{\prime }-10 \,{\mathrm e}^{x +y}=x^{2}} \]
Writing the ode as \begin {align*} y^{\prime } &= 10 \,{\mathrm e}^{x +y}+x^{2}\tag {1} \end {align*}
And using the substitution \(u={\mathrm e}^{-y}\) then \begin {align*} u' &= -y^{\prime } {\mathrm e}^{-y} \end {align*}
The above shows that \begin {align*} y^{\prime } &= -u^{\prime }\left (x \right ) {\mathrm e}^{y}\\ &= -\frac {u^{\prime }\left (x \right )}{u} \end {align*}
Substituting this in (1) gives \begin {align*} -\frac {u^{\prime }\left (x \right )}{u}&=\frac {10 \,{\mathrm e}^{x}}{u}+x^{2} \end {align*}
The above simplifies to \begin {align*} -u^{\prime }\left (x \right )&=10 \,{\mathrm e}^{x}+x^{2} u \left (x \right )\\ u^{\prime }\left (x \right )+x^{2} u \left (x \right )&=-10 \,{\mathrm e}^{x}\tag {2} \end {align*}
Now ode (2) is solved for \(u \left (x \right )\)
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} u^{\prime }\left (x \right ) + p(x)u \left (x \right ) &= q(x) \end {align*}
Where here \begin {align*} p(x) &=x^{2}\\ q(x) &=-10 \,{\mathrm e}^{x} \end {align*}
Hence the ode is \begin {align*} u^{\prime }\left (x \right )+x^{2} u \left (x \right ) = -10 \,{\mathrm e}^{x} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int x^{2}d x} \\ &= {\mathrm e}^{\frac {x^{3}}{3}} \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu u\right ) &= \left (\mu \right ) \left (-10 \,{\mathrm e}^{x}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left ({\mathrm e}^{\frac {x^{3}}{3}} u\right ) &= \left ({\mathrm e}^{\frac {x^{3}}{3}}\right ) \left (-10 \,{\mathrm e}^{x}\right )\\ \mathrm {d} \left ({\mathrm e}^{\frac {x^{3}}{3}} u\right ) &= \left (-10 \,{\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} {\mathrm e}^{\frac {x^{3}}{3}} u &= \int {-10 \,{\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}\,\mathrm {d} x}\\ {\mathrm e}^{\frac {x^{3}}{3}} u &= \int -10 \,{\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}d x + c_{1} \end {align*}
Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\frac {x^{3}}{3}}\) results in \begin {align*} u \left (x \right ) &= {\mathrm e}^{-\frac {x^{3}}{3}} \left (\int -10 \,{\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}d x \right )+c_{1} {\mathrm e}^{-\frac {x^{3}}{3}} \end {align*}
which simplifies to \begin {align*} u \left (x \right ) &= {\mathrm e}^{-\frac {x^{3}}{3}} \left (-10 \left (\int {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}d x \right )+c_{1} \right ) \end {align*}
Substituting the solution found for \(u \left (x \right )\) in \(u={\mathrm e}^{-y}\) gives \begin {align*} y&= -\ln \left (u \left (x \right )\right )\\ &= -\ln \left ({\mathrm e}^{-\frac {x^{3}}{3}} \left (-10 \left (\int {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}d x \right )+c_{1} \right )\right )\\ &= -\ln \left ({\mathrm e}^{-\frac {x^{3}}{3}} \left (-10 \left (\int {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}d x \right )+c_{1} \right )\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\ln \left ({\mathrm e}^{-\frac {x^{3}}{3}} \left (-10 \left (\int {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}d x \right )+c_{1} \right )\right ) \\ \end{align*}
Verification of solutions
\[ y = -\ln \left ({\mathrm e}^{-\frac {x^{3}}{3}} \left (-10 \left (\int {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}d x \right )+c_{1} \right )\right ) \] Verified OK.
Writing the ode as \begin {align*} y^{\prime }&=10 \,{\mathrm e}^{x +y}+x^{2}\\ y^{\prime }&= \omega \left ( x,y\right ) \end {align*}
The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end {align*}
The type of this ode is known. It is of type first order special form ID 1
. 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 |
\(y^{\prime }=f_{1}\left ( x\right ) y+f_{2}\left ( x\right )y^{2}\) |
\(0\) |
\(e^{-\int f_{1}dx}\) |
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The above table shows that \begin {align*} \xi \left (x,y\right ) &=\frac {{\mathrm e}^{-\frac {1}{3} x^{3}-x}}{10}\\ \tag {A1} \eta \left (x,y\right ) &=x^{2}+\frac {{\mathrm e}^{-\frac {1}{3} x^{3}-x}}{10} \end {align*}
The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) 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 y}{\eta } = dS \tag {1} \end {align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). 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 \(S(R)\). Therefore \begin {align*} \frac {dy}{dx} &= \frac {\eta }{\xi }\\ &= \frac {x^{2}+\frac {{\mathrm e}^{-\frac {1}{3} x^{3}-x}}{10}}{\frac {{\mathrm e}^{-\frac {1}{3} x^{3}-x}}{10}}\\ &= 10 \,{\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}} x^{2}+1 \end {align*}
This is easily solved to give \begin {align*} y = \int \left (10 \,{\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}} x^{2}+1\right )d x +c_{1} \end {align*}
Where now the coordinate \(R\) is taken as the constant of integration. Hence \begin {align*} R &= -\left (\int \left (10 \,{\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}} x^{2}+1\right )d x \right )+y \end {align*}
And \(S\) is found from \begin {align*} dS &= \frac {dx}{\xi } \\ &= \frac {dx}{\frac {{\mathrm e}^{-\frac {1}{3} x^{3}-x}}{10}} \end {align*}
Integrating gives \begin {align*} S &= \int { \frac {dx}{T}}\\ &= \int 10 \,{\mathrm e}^{\frac {1}{3} x^{3}+x}d x \end {align*}
Where the constant of integration is set to zero as we just need one solution.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-10 \,{\mathrm e}^{x +y}=x^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=10 \,{\mathrm e}^{x +y}+x^{2} \end {array} \]
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 Looking for potential symmetries trying inverse_Riccati trying an equivalence to an Abel ODE differential order: 1; trying a linearization to 2nd order --- trying a change of variables {x -> y(x), y(x) -> x} differential order: 1; trying a linearization to 2nd order trying 1st order ODE linearizable_by_differentiation --- Trying Lie symmetry methods, 1st order --- `, `-> Computing symmetries using: way = 3 `, `-> Computing symmetries using: way = 4 `, `-> Computing symmetries using: way = 5 trying symmetry patterns for 1st order ODEs -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] <- symmetry pattern of the form [0, F(x)*G(y)] successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 30
dsolve(diff(y(x),x)=10*exp(x+y(x))+x^2,y(x), singsol=all)
\[ y \left (x \right ) = \frac {x^{3}}{3}-\ln \left (-c_{1} -10 \left (\int {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}d x \right )\right ) \]
✓ Solution by Mathematica
Time used: 0.431 (sec). Leaf size: 115
DSolve[y'[x]==10*Exp[x+y[x]]+x^2,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^{y(x)}-\frac {1}{10} e^{-K[2]} \left (10 e^{K[2]} \int _1^x-\frac {1}{10} e^{\frac {K[1]^3}{3}-K[2]} K[1]^2dK[1]+e^{\frac {x^3}{3}}\right )dK[2]+\int _1^x\left (\frac {1}{10} e^{\frac {K[1]^3}{3}-y(x)} K[1]^2+e^{\frac {K[1]^3}{3}+K[1]}\right )dK[1]=c_1,y(x)\right ] \]