2.3.29 problem 29

Internal problem ID [8563]
Book : Own collection of miscellaneous problems
Section : section 3.0
Problem number : 29
Date solved : Tuesday, December 17, 2024 at 12:56:03 PM
CAS classification : [[_homogeneous, `class D`]]

Solve

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

Solved as first order homogeneous class D ode

Time used: 0.233 (sec)

Writing the ode as

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

The given ode has the form

\begin{equation} y^{\prime }=\frac {y}{x}+g\left ( x\right ) f\left ( b\frac {y}{x}\right ) ^{\frac {n}{m}}\tag {1}\end{equation}

Where \(b\) is scalar and \(g\left ( x\right ) \) is function of \(x\) and \(n,m\) are integers. The solution is given in Kamke page 20. Using the substitution \(y\left ( x\right ) =u\left ( x\right ) x\) then

\[ \frac {dy}{dx}=\frac {du}{dx}x+u \]

Hence the given ode becomes

\begin{align} \frac {du}{dx}x+u & =u+g\left ( x\right ) f\left ( bu\right ) ^{\frac {n}{m}}\nonumber \\ u^{\prime } & =\frac {1}{x}g\left ( x\right ) f\left ( bu\right ) ^{\frac {n}{m}}\tag {2}\end{align}

The above ode is always separable. This is easily solved for \(u\) assuming the integration can be resolved, and then the solution to the original ode becomes \(y=ux\). Comapring the given ode (A) with the form (1) shows that

\begin{align*} g \left (x \right )&=2 x^{2}\\ b&=1\\ f \left (\frac {b x}{y}\right )&=\sin \left (\frac {y}{x}\right ) \end{align*}

Substituting the above in (2) results in the \(u(x)\) ode as

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

Which is now solved as separable The ode \(u^{\prime }\left (x \right ) = 2 x \sin \left (u \left (x \right )\right )^{2}\) is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= 2 x \sin \left (u \left (x \right )\right )^{2}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= 2 x\\ g(u) &= \sin \left (u \right )^{2} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx}\\ \int { \frac {1}{\sin \left (u \right )^{2}}\,du} &= \int { 2 x \,dx}\\ -\cot \left (u \left (x \right )\right )&=x^{2}+c_1 \end{align*}

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or \(\sin \left (u \right )^{2}=0\) for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right )&=0 \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} -\cot \left (u \left (x \right )\right ) = x^{2}+c_1\\ u \left (x \right ) = 0 \end{align*}

Solving for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right ) &= 0 \\ u \left (x \right ) &= \pi -\operatorname {arccot}\left (x^{2}+c_1 \right ) \\ \end{align*}

Converting \(u \left (x \right ) = 0\) back to \(y\) gives

\begin{align*} y = 0 \end{align*}

Converting \(u \left (x \right ) = \pi -\operatorname {arccot}\left (x^{2}+c_1 \right )\) back to \(y\) gives

\begin{align*} y = x \left (\pi -\operatorname {arccot}\left (x^{2}+c_1 \right )\right ) \end{align*}

Figure 2.215: Slope field plot
\(y^{\prime } = 2 x^{2} \sin \left (\frac {y}{x}\right )^{2}+\frac {y}{x}\)

Summary of solutions found

\begin{align*} y &= 0 \\ y &= x \left (\pi -\operatorname {arccot}\left (x^{2}+c_1 \right )\right ) \\ \end{align*}

Solved as first order homogeneous class D2 ode

Time used: 0.077 (sec)

Applying change of variables \(y = u \left (x \right ) x\), then the ode becomes

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

Which is now solved The ode \(u^{\prime }\left (x \right ) = 2 x \sin \left (u \left (x \right )\right )^{2}\) is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= 2 x \sin \left (u \left (x \right )\right )^{2}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= 2 x\\ g(u) &= \sin \left (u \right )^{2} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx}\\ \int { \frac {1}{\sin \left (u \right )^{2}}\,du} &= \int { 2 x \,dx}\\ -\cot \left (u \left (x \right )\right )&=x^{2}+c_1 \end{align*}

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or \(\sin \left (u \right )^{2}=0\) for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right )&=0 \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} -\cot \left (u \left (x \right )\right ) = x^{2}+c_1\\ u \left (x \right ) = 0 \end{align*}

Solving for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right ) &= 0 \\ u \left (x \right ) &= \pi -\operatorname {arccot}\left (x^{2}+c_1 \right ) \\ \end{align*}

Converting \(u \left (x \right ) = 0\) back to \(y\) gives

\begin{align*} y = 0 \end{align*}

Converting \(u \left (x \right ) = \pi -\operatorname {arccot}\left (x^{2}+c_1 \right )\) back to \(y\) gives

\begin{align*} y = x \left (\pi -\operatorname {arccot}\left (x^{2}+c_1 \right )\right ) \end{align*}

Figure 2.216: Slope field plot
\(y^{\prime } = 2 x^{2} \sin \left (\frac {y}{x}\right )^{2}+\frac {y}{x}\)

Summary of solutions found

\begin{align*} y &= 0 \\ y &= x \left (\pi -\operatorname {arccot}\left (x^{2}+c_1 \right )\right ) \\ \end{align*}

Maple step by step solution

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

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 homogeneous D 
<- homogeneous successful`
 

Maple dsolve solution

Solving time : 0.008 (sec)
Leaf size : 18

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

\[ y = \left (\frac {\pi }{2}+\arctan \left (x^{2}+2 c_{1} \right )\right ) x \]

Mathematica DSolve solution

Solving time : 0.307 (sec)
Leaf size : 22

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

\begin{align*} y(x)\to -x \cot ^{-1}\left (x^2-2 c_1\right ) \\ y(x)\to 0 \\ \end{align*}