1.27 problem 27

Internal problem ID [7991]
Book : First order enumerated odes
Section : section 1
Problem number : 27
Date solved : Monday, October 21, 2024 at 04:40:30 PM
CAS classification : [_Riccati]

Solve

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

1.27.1 Solved as first order ode of type Riccati

Time used: 0.461 (sec)

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= \sin \left (x \right )+y^{2} \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = \sin \left (x \right )+y^{2} \]

With Riccati ODE standard form

\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]

Shows that \(f_0(x)=\sin \left (x \right )\), \(f_1(x)=0\) and \(f_2(x)=1\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u} \tag {1} \end{align*}

Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for \(u(x)\) which is

\begin{align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end{align*}

But

\begin{align*} f_2' &=0\\ f_1 f_2 &=0\\ f_2^2 f_0 &=\sin \left (x \right ) \end{align*}

Substituting the above terms back in equation (2) gives

\begin{align*} u^{\prime \prime }\left (x \right )+\sin \left (x \right ) u \left (x \right ) = 0 \end{align*}

Solution obtained is

\[ u \left (x \right ) = c_1 \operatorname {MathieuC}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )+c_2 \operatorname {MathieuS}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right ) \]

Taking derivative gives

\[ u^{\prime }\left (x \right ) = \frac {c_1 \operatorname {MathieuCPrime}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )}{2}+\frac {c_2 \operatorname {MathieuSPrime}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )}{2} \]

Doing change of constants, the solution becomes

\[ y = -\frac {\frac {c_1 \operatorname {MathieuCPrime}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )}{2}+\frac {\operatorname {MathieuSPrime}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )}{2}}{c_1 \operatorname {MathieuC}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )+\operatorname {MathieuS}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )} \]

Figure 23: Slope field plot
\(y^{\prime } = \sin \left (x \right )+y^{2}\)

1.27.2 Maple step by step solution

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

1.27.3 Maple trace

Methods for first order ODEs:
 

1.27.4 Maple dsolve solution

Solving time : 0.003 (sec)
Leaf size : 59

dsolve(diff(y(x),x) = sin(x)+y(x)^2, 
       y(x),singsol=all)
 

\[ y = \frac {-c_1 \operatorname {MathieuSPrime}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )-\operatorname {MathieuCPrime}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )}{2 c_1 \operatorname {MathieuS}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )+2 \operatorname {MathieuC}\left (0, -2, -\frac {\pi }{4}+\frac {x}{2}\right )} \]

1.27.5 Mathematica DSolve solution

Solving time : 0.174 (sec)
Leaf size : 105

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

\begin{align*} y(x)\to \frac {-\text {MathieuSPrime}\left [0,-2,\frac {1}{4} (\pi -2 x)\right ]+c_1 \text {MathieuCPrime}\left [0,-2,\frac {1}{4} (\pi -2 x)\right ]}{2 \left (\text {MathieuS}\left [0,-2,\frac {1}{4} (2 x-\pi )\right ]+c_1 \text {MathieuC}\left [0,-2,\frac {1}{4} (\pi -2 x)\right ]\right )} \\ y(x)\to \frac {\text {MathieuCPrime}\left [0,-2,\frac {1}{4} (\pi -2 x)\right ]}{2 \text {MathieuC}\left [0,-2,\frac {1}{4} (\pi -2 x)\right ]} \\ \end{align*}