Question

Solve the differential equation x^(3)p^(2)+x^(2)yp+a^(3)=0 and also obtain its singular solution if

Answer 1

To begin with, we clarify that by expression $p$ we mean the first derivative with respect to the variable $x$ :

$$p=\frac{dy}{dx}$$

Now we rewrite the initial equation, expressing the variable $y$ as a function of $x$ and $p$

$$x^3p^2+x^2yp+a^3=0\longrightarrow\\[0.3cm] \left. x^2yp=-a^3-x^3p^2\right|\div\left(x^2p\right)\longrightarrow\\[0.3cm] y=-\frac{a^3}{x^2p}-\frac{x^3p^2}{x^2p}\longrightarrow\\[0.3cm]\boxed{y=-\frac{a^3}{x^2p}-xp}$$

We differentiate the resulting expression with respect to $x$ , remembering that $p=dy/dx$ in order to exclude the variable $y$ from the equation.

$$\frac{d}{dx}\left|y=-\frac{a^3}{x^2p}-xp\right.\longrightarrow\\[0.3cm] p=-p-x\cdot\frac{dp}{dx}-\frac{a^3}{x^3p}\cdot(-2)-\frac{a^3}{x^2p^2}\cdot\left(-\frac{dp}{dx}\right)\\[0.3cm] p+p-\frac{2a^3}{x^3p}=-\left(x-\frac{a^3}{x^2p^2}\right)\cdot\frac{dp}{dx}\\[0.3cm] 2p\cdot\left(1-\frac{a^3}{x^3p^2}\right)=-x\cdot\left(1-\frac{a^3}{x^3p^2}\right)\cdot\frac{dp}{dx}\\[0.3cm] 2p=-x\cdot\displaystyle\frac{dp}{dx}$$

$$2p=-x\cdot\frac{dp}{dx}\longrightarrow\frac{dp}{p}=-\frac{dx}{2x}\longrightarrow\\[0.3cm] \ln|p|=-2\cdot\ln|x|+\ln|C|\longrightarrow\\[0.3cm] \ln|p|+\ln\left|x^2\right|=\ln|C|\longrightarrow px^2=C\\[0.3cm] \boxed{p=\frac{C}{x^2}}$$

Substitute the resulting ratio in the original equation

$$x^3p^2+x^2yp+a^3=0\longrightarrow\\[0.3cm] x^3\cdot\left(\frac{C}{x^2}\right)^2+x^2y\cdot\left(\frac{C}{x^2}\right)+a^3=0\longrightarrow\\[0.3cm] \left.\frac{C^2}{x}+Cy+a^3=0\right|\times(x)\\[0.3cm] \boxed{C^2+(xy)C+a^3x=0}$$

which is a quadratic equation in $C$ . Equating it's discriminant to 0, we get :

$$D=x^2y^2-4a^3x=0\longrightarrow\\[0.3cm] x\cdot\left(xy^2-4a^3\right)=0\longrightarrow \left[\begin{array}{l} x=0\\[0.3cm] xy^2-4a^3=0 \end{array}\right.$$

ANSWER

$$\left[\begin{array}{l} x=0\\[0.3cm] xy^2-4a^3=0 \end{array}\right.$$

are the singular solutions of the given differential equation.