Find the integral curve of y y'' + (y')^2 = 0 which passes…

Find the integral curve of $y\,y'' + (y')^2 = 0$ which passes through the origin and whose tangent slope at any point equals the reciprocal of the ordinate at that point.

Step 1: Solve the differential equation

The left-hand side is an exact derivative: $y\,y'' + (y')^2 = \frac{d}{dx}(y\,y').$

Thus the equation becomes $\dfrac{d}{dx}(y\,y')=0$. Integrate: \[y\,y' = C_1. \tag{1}\]

Step 2: Second integration

Separate variables in (1) and integrate: \[y\,dy = C_1\,dx,\qquad \frac{y^2}{2} = C_1\,x + C_2,\qquad y^2 = 2C_1\,x + 2C_2. \tag{2}\]

The general solution is $y^2 = Ax+B$ with $A=2C_1$, $B=2C_2$ arbitrary.

Step 3: Apply the initial conditions

Through the origin: substitute $x=0,\,y=0$ into (2): $0 = B \implies B = 0$, so $y^2 = Ax$.

Slope condition $y' = 1/y$: from (1) with $B=0$, $y\,y' = C_1$, and since $y' = 1/y$, $y\cdot\frac1y = C_1 \implies C_1 = 1$, so $A = 2$.

Step 4: The particular integral curve

The particular integral curve is recorded in the result below.

Step 5: Verification

Passes through origin: $0^2 = 2\cdot 0 = 0$. ✓

Slope condition: differentiate $y^2=2x$: $2yy'=2 \implies y'=1/y$. ✓

Satisfies the ODE: from $y'=1/y$, $y'' = -y'/y^2 = -1/y^3$. Check $yy''+(y')^2$:

\[ y\cdot\left(-\frac{1}{y^3}\right) + \frac{1}{y^2} = -\frac{1}{y^2}+\frac{1}{y^2} = 0.\ \checkmark \]