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`]]
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
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*}
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*}
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
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*}