Discuss the solvability of the differential equation \[ (e^x\sin y)(y')^3+(e^x\cos y)y'+e^y\tan x=0 \] with the initial condition \(y(0)=0\). Does a solution exist in some interval about \(0\)? If so, is it unique?
Solution
Define \[ F(x,y,p) = e^x\sin y\,p^3+e^x\cos y\,p+e^y\tan x. \]
1. Any classical solution with \(y(0)=0\) must satisfy\[ y'(0)=0. \]
At \(x=0\) and \(y=0\), \[ F(0,0,p)=p. \] Thus the differential equation \(F(x,y,y')=0\) at the initial point gives \[ 0=F(0,0,y'(0))=y'(0). \]
2. Near \((x,y,p)=(0,0,0)\), the equation \(F(x,y,p)=0\) is equivalent to an explicit equation\[ p=\phi(x,y) \]for a uniquely determined smooth function \(\phi\) with \(\phi(0,0)=0\).
The function \(F\) is smooth in a neighborhood of \((0,0,0)\), since \(\tan x\) is smooth near \(x=0\). Moreover, \[ \frac{\partial F}{\partial p}(x,y,p) = 3e^x\sin y\,p^2+e^x\cos y, \] so \[ \frac{\partial F}{\partial p}(0,0,0)=1\neq0. \] The implicit-function theorem therefore gives neighborhoods of \((0,0)\) and \(0\) and a unique smooth function \(\phi\) on the first neighborhood such that, within these neighborhoods, \[ F(x,y,p)=0 \quad\Longleftrightarrow\quad p=\phi(x,y). \] Since \(F(0,0,0)=0\), one has \(\phi(0,0)=0\).
3. There exists a solution of the original differential equation on some interval about \(0\) with \(y(0)=0\).
By step 2, it is enough near the initial point to solve \[ y'=\phi(x,y), \qquad y(0)=0. \] The function \(\phi\) is smooth, hence continuous and locally Lipschitz in \(y\). The local existence theorem for first-order ordinary differential equations therefore gives a solution on some interval \((-\varepsilon,\varepsilon)\) about \(0\). By construction, this solution also satisfies \(F(x,y,y')=0\), which is the original equation.
4. The solution in step 3 is locally unique among classical solutions satisfying \(y(0)=0\).
Let \(\widetilde y\) be any classical solution of the original equation with \(\widetilde y(0)=0\). By step 1, \[ \widetilde y'(0)=0. \] Since \(\widetilde y\) and \(\widetilde y'\) are continuous, after shrinking the interval if necessary the triple \[ (x,\widetilde y(x),\widetilde y'(x)) \] stays in the neighborhood where step 2 applies. Hence \[ \widetilde y'(x)=\phi(x,\widetilde y(x)). \] The local uniqueness theorem for \[ y'=\phi(x,y), \qquad y(0)=0 \] then implies that \(\widetilde y\) agrees with the solution from step 3 on a neighborhood of \(0\).
5. Therefore the initial-value problem has\[ \boxed{\text{a locally existing and locally unique solution about }0}. \]
Existence is step 3 and uniqueness is step 4.
6. Q.E.D.
Step 5 answers both questions in the problem.