3.29 problem 29

3.29.1 Solved as first order homogeneous class D ode
3.29.2 Solved as first order homogeneous class D2 ode
3.29.3 Maple step by step solution
3.29.4 Maple trace
3.29.5 Maple dsolve solution
3.29.6 Mathematica DSolve solution

Internal problem ID [7867]
Book : Own collection of miscellaneous problems
Section : section 3.0
Problem number : 29
Date solved : Monday, October 21, 2024 at 04:30:28 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*}

3.29.1 Solved as first order homogeneous class D ode

Time used: 0.164 (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 \(u \left (x \right )\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\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 205: Slope field plot
\(y^{\prime } = 2 x^{2} \sin \left (\frac {y}{x}\right )^{2}+\frac {y}{x}\)
3.29.2 Solved as first order homogeneous class D2 ode

Time used: 0.065 (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 \(u \left (x \right )\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\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 206: Slope field plot
\(y^{\prime } = 2 x^{2} \sin \left (\frac {y}{x}\right )^{2}+\frac {y}{x}\)
3.29.3 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }=2 x^{2} \sin \left (\frac {y}{x}\right )^{2}+\frac {y}{x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=2 x^{2} \sin \left (\frac {y}{x}\right )^{2}+\frac {y}{x} \end {array} \]

3.29.4 Maple trace
Methods for first order ODEs:
 
3.29.5 Maple dsolve solution

Solving time : 0.003 (sec)
Leaf size : 18

dsolve(diff(y(x),x) = 2*x^2*sin(y(x)/x)^2+y(x)/x, 
       y(x),singsol=all)
 
\[ y = \left (\frac {\pi }{2}+\arctan \left (x^{2}+2 c_1 \right )\right ) x \]
3.29.6 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*}