2.1.70 problem 70

Internal problem ID [8730]
Book : First order enumerated odes
Section : section 1
Problem number : 70
Date solved : Tuesday, December 17, 2024 at 01:01:46 PM
CAS classification : [[_1st_order, `_with_symmetry_[F(x),G(x)]`]]

Solve

\begin{align*} y^{\prime }&=5 \,{\mathrm e}^{x^{2}+20 y}+\sin \left (x \right ) \end{align*}

Solved as first order ode of type ID 1

Time used: 0.685 (sec)

Writing the ode as

\begin{align*} y^{\prime } &= 5 \,{\mathrm e}^{x^{2}+20 y}+\sin \left (x \right )\tag {1} \end{align*}

And using the substitution \(u={\mathrm e}^{-20 y}\) then

\begin{align*} u' &= -20 y^{\prime } {\mathrm e}^{-20 y} \end{align*}

The above shows that

\begin{align*} y^{\prime } &= -\frac {u^{\prime }\left (x \right ) {\mathrm e}^{20 y}}{20}\\ &= -\frac {u^{\prime }\left (x \right )}{20 u} \end{align*}

Substituting this in (1) gives

\begin{align*} -\frac {u^{\prime }\left (x \right )}{20 u}&=\frac {5 \,{\mathrm e}^{x^{2}}}{u}+\sin \left (x \right ) \end{align*}

The above simplifies to

\begin{align*} -\frac {u^{\prime }\left (x \right )}{20}&=5 \,{\mathrm e}^{x^{2}}+\sin \left (x \right ) u \left (x \right )\\ u^{\prime }\left (x \right )+20 \sin \left (x \right ) u \left (x \right )&=-100 \,{\mathrm e}^{x^{2}}\tag {2} \end{align*}

Now ode (2) is solved for \(u \left (x \right )\).

In canonical form a linear first order is

\begin{align*} u^{\prime }\left (x \right ) + q(x)u \left (x \right ) &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=20 \sin \left (x \right )\\ p(x) &=-100 \,{\mathrm e}^{x^{2}} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int 20 \sin \left (x \right )d x}\\ &= {\mathrm e}^{-20 \cos \left (x \right )} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu u\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu u\right ) &= \left (\mu \right ) \left (-100 \,{\mathrm e}^{x^{2}}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (u \,{\mathrm e}^{-20 \cos \left (x \right )}\right ) &= \left ({\mathrm e}^{-20 \cos \left (x \right )}\right ) \left (-100 \,{\mathrm e}^{x^{2}}\right ) \\ \mathrm {d} \left (u \,{\mathrm e}^{-20 \cos \left (x \right )}\right ) &= \left (-100 \,{\mathrm e}^{x^{2}} {\mathrm e}^{-20 \cos \left (x \right )}\right )\, \mathrm {d} x \\ \end{align*}

Integrating gives

\begin{align*} u \,{\mathrm e}^{-20 \cos \left (x \right )}&= \int {-100 \,{\mathrm e}^{x^{2}} {\mathrm e}^{-20 \cos \left (x \right )} \,dx} \\ &=\int -100 \,{\mathrm e}^{x^{2}} {\mathrm e}^{-20 \cos \left (x \right )}d x + c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{-20 \cos \left (x \right )}\) gives the final solution

\[ u \left (x \right ) = {\mathrm e}^{20 \cos \left (x \right )} \left (\int -100 \,{\mathrm e}^{x^{2}} {\mathrm e}^{-20 \cos \left (x \right )}d x +c_1 \right ) \]

Substituting the solution found for \(u \left (x \right )\) in \(u={\mathrm e}^{-20 y}\) gives

\begin{align*} y&= -\frac {\ln \left (u \left (x \right )\right )}{20}\\ &= -\frac {\ln \left (-\frac {\ln \left (\left (-100 \left (\int {\mathrm e}^{x^{2}-20 \cos \left (x \right )}d x \right )+c_1 \right ) {\mathrm e}^{20 \cos \left (x \right )}\right )}{20}\right )}{20}\\ &= -\frac {\ln \left (\left (-100 \left (\int {\mathrm e}^{x^{2}-20 \cos \left (x \right )}d x \right )+c_1 \right ) {\mathrm e}^{20 \cos \left (x \right )}\right )}{20} \end{align*}

Figure 2.99: Slope field plot
\(y^{\prime } = 5 \,{\mathrm e}^{x^{2}+20 y}+\sin \left (x \right )\)

Summary of solutions found

\begin{align*} y &= -\frac {\ln \left (\left (-100 \left (\int {\mathrm e}^{x^{2}-20 \cos \left (x \right )}d x \right )+c_1 \right ) {\mathrm e}^{20 \cos \left (x \right )}\right )}{20} \\ \end{align*}

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=5 \,{\mathrm e}^{x^{2}+20 y \left (x \right )}+\sin \left (x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=5 \,{\mathrm e}^{x^{2}+20 y \left (x \right )}+\sin \left (x \right ) \end {array} \]

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 
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`
 

Maple dsolve solution

Solving time : 0.023 (sec)
Leaf size : 33

dsolve(diff(y(x),x) = 5*exp(x^2+20*y(x))+sin(x), 
       y(x),singsol=all)
 

\[ y = -\cos \left (x \right )-\frac {\ln \left (20\right )}{20}-\frac {\ln \left (-c_{1} -5 \left (\int {\mathrm e}^{x^{2}-20 \cos \left (x \right )}d x \right )\right )}{20} \]

Mathematica DSolve solution

Solving time : 7.542 (sec)
Leaf size : 140

DSolve[{D[y[x],x]==5*Exp[x^2+20*y[x]]+Sin[x],{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ \text {Solve}\left [\int _1^x-\frac {1}{100} e^{-20 \cos (K[1])-20 y(x)} \left (\sin (K[1])+5 e^{K[1]^2+20 y(x)}\right )dK[1]+\int _1^{y(x)}-\frac {1}{100} e^{-20 \cos (x)-20 K[2]} \left (100 e^{20 \cos (x)+20 K[2]} \int _1^x\left (\frac {1}{5} e^{-20 \cos (K[1])-20 K[2]} \left (\sin (K[1])+5 e^{K[1]^2+20 K[2]}\right )-e^{K[1]^2-20 \cos (K[1])}\right )dK[1]-1\right )dK[2]=c_1,y(x)\right ] \]