Find the general solution of xyy'' + x(y')^2 = yy' , then…

Find the general solution of $xyy'' + x(y')^2 = yy'$, then find the integral curve passing through both $(2,0)$ and $(0,1)$, and compute the area it encloses.

Step 1: Recognizing a total derivative

The combination $yy''+(y')^2$ is exactly the derivative of $yy'$: $\frac{d}{dx}(yy') = yy'' + (y')^2$.

The left side of the given equation, $x\bigl[yy''+(y')^2\bigr]$, is therefore $x\,\dfrac{d}{dx}(yy')$, while the right side is $yy'$ itself. Setting $u=yy'$, the equation $xyy''+x(y')^2=yy'$ becomes simply

\[ x\frac{du}{dx} = u. \]

Step 2: Solving for $u$

This is separable: \[\frac{du}{u} = \frac{dx}{x} \;\Longrightarrow\; \ln|u| = \ln|x| + \text{const.} \;\Longrightarrow\; u = ax,\]

for an arbitrary constant $a$. Recalling $u=yy'$, this gives the first-order equation $yy'=ax$.

Step 3: Solving for $y$

This is separable in $y$ and $x$: $y\,dy = ax\,dx$. Integrating, \[\frac{y^2}{2} = \frac{ax^2}{2} + \frac{b}{2} \;\Longrightarrow\; y^2 = ax^2 + b,\]

for arbitrary constants $a,b$ — a two-parameter family of central conics (ellipses or hyperbolas, depending on the sign of $a$ relative to $b$) with axes along the coordinate axes.

Step 4: Verification of the general solution

Differentiating $y^2=ax^2+b$ implicitly gives $2yy'=2ax$, i.e. $(yy')'=a$ after one more differentiation (since $2yy'=2ax$ differentiates to $2(yy')'=2a$). Then

\[ xyy''+x(y')^2 = x(yy')' = x\cdot a = ax = yy', \]

which is exactly the original equation, confirmed identically for every $a,b$. $\checkmark$

Step 5: The curve through $(2,0)$ and $(0,1)$

Impose the two point conditions on $y^2=ax^2+b$. At $(0,1)$: $1=a\cdot0+b$, so $b=1$. At $(2,0)$: $0=4a+b=4a+1$, so $a=-\tfrac14$. The specific curve is

\[ y^2 = -\frac{x^2}{4} + 1, \qquad \text{i.e.} \qquad \frac{x^2}{4} + y^2 = 1, \]

an ellipse centered at the origin with semi-axes $2$ (along $x$) and $1$ (along $y$). Checking: at $x=2$, $y^2=1-1=0\ \checkmark$; at $x=0$, $y^2=1\ \checkmark$.

Step 6: Enclosed area

For an ellipse $x^2/A^2+y^2/B^2=1$ with semi-axes $A,B$, the enclosed area is $S=\pi AB$. Here $A=2$, $B=1$, so

\[ S = \pi\cdot 2\cdot 1 = 2\pi. \]

As a cross-check, compute the area directly by integration. Solving for the upper half, $y=\sqrt{1-x^2/4}$, the total area is

\[ S = 2\int_{-2}^{2}\sqrt{1-\frac{x^2}{4}}\,dx. \]

Substitute $x=2\sin\theta$, $dx=2\cos\theta\,d\theta$, with $\theta$ running from $-\pi/2$ to $\pi/2$: \[S = 2\int_{-\pi/2}^{\pi/2}\sqrt{1-\sin^2\theta}\cdot 2\cos\theta\,d\theta = 4\int_{-\pi/2}^{\pi/2}\cos^2\theta\,d\theta = 4\cdot\frac{\pi}{2} = 2\pi,\]

confirming the ellipse-area formula result exactly.