# Help with a first-order implicit differential equation

$$2xp=2\tan y+p^{3}\cos^{2}y$$ ($$p=\frac{dy}{dx}$$)

Multiply both sides by $$\cos y$$, $$2xp\cos y=2\sin y+p^{3}\cos ^{3}y$$

Let $$t=p\cos y$$,the equation becomes$$\sin y=xt-\frac{1}{2}t^{3}$$ Take the derivative of both sides with respect to $$x$$, $$t=t+x \frac{dt}{dx}-\frac{3}{2}t^{2} \frac{dt}{dx}$$ $$\left( x-\frac{3}{2}t^{2} \right) \frac{dt}{dx}=0$$ I don't know what to do next,$$x- \frac{3}{2}t^{2}=0$$ and $$\frac{dt}{dx}=0$$ still equation in terms of $$x,y$$ and $$\frac{dy}{dx}(p)$$.

Your transformations are very suggestive leading to consider $$z=\sin y$$. Then the equation becomes, with $$z'=\cos(y)y'=t$$, $$z=xz'-\frac12(z')^3=xt-\frac12t^3.$$ This now has the form of a Clairaut DE, with its immediate solution by lines and the envelope.
• $z=xz'-\frac12\,(z')^3$ is scale-invariant under $x\rightarrow \alpha x, z\rightarrow \alpha^{3/2}z$, so it can be further reduced to an equidimensional-in-$x$ equation by substituting $z=x^{3/2}u(x)$. Commented Aug 6 at 13:51
• This is making the solution more complicated. From Clairaut you get directly the solutions $z=Cx-\frac12C^3$. And the singular curve where $x=\frac32(z')^2$ which can be inserted into $z=z'(x-\frac12(z')^2)=\pm\frac23x\sqrt{\frac23x}$. Commented Aug 6 at 17:06
Let's first solve $$\frac{dt}{dx}=0$$: $$\frac{dt}{dx}=0 \implies t=C_1 \implies \frac{dy}{dx}\cos y= C_1 \implies \sin y=C_1x+C_2. \tag{1}$$ Since the original ODE is of first order, we expect that its general solution has a single constant of integration, whereas $$(1)$$ has two. To determine the relation beween $$C_1$$ and $$C_2$$, we substitute $$(1)$$ in the original ODE: \begin{align} 2x\frac{C_1}{\cos y}=2\tan y+\frac{C_1^3}{\cos^3 y}\cos^2y \implies 2C_1x=2\sin y+C_1^3 \\ \implies 2C_1x=2(C_1x+C_2)+C_1^3 \implies C_2=-\frac{C_1^3}{2}. \tag{2} \end{align} Therefore, the general solution to the original ODE is $$\sin y=C_1x-\frac{C_1^3}{2}. \tag{3}$$
Next, instead of solving $$x-\frac{3}{2}t^2=0$$, we notice that the envelope of $$(3)$$ is also a solution to the original ODE. To find it, eliminate $$C_1$$ between $$(3)$$ and $$\frac{\partial}{\partial C_1}\left\{\sin y-C_1x+\frac{C_1^3}{2}\right\}=0 \implies x-\frac{3C_1^2}{2}=0. \tag{4}$$ Solving $$(4)$$ for $$C_1$$ and plugging the result into $$(3)$$, we obtain $$\sin y=\pm\left(\frac{2x}{3}\right)^{3/2}\qquad\left(0\leq x\leq \frac{3}{2}\right). \tag{5}$$ It's straightforward to show that $$(5)$$ is a solution to $$x-\frac{3}{2}t^2=0$$.