Find the curves for which the tangent at each point forms…
Find the curves for which the tangent at each point forms with the coordinate axes and the ordinate of the point of tangency a trapezoid of constant area $a^2$. Among these curves, find the one through the point $(a,a)$.
Explanation of Method
Express the trapezoid's area in terms of $x,y,y'$, set it equal to $a^2$, and solve the resulting linear equation; then apply the initial condition.
Step 1: Set up the area condition
The trapezoid has parallel sides $y-xy'$ (on the $y$-axis) and $y$ (the ordinate at $x$), separated by distance $x$, so its area is $\dfrac12\big[(y-xy')+y\big]x=xy-\dfrac{x^2y'}{2}$. Setting this equal to $a^2$:
\[2xy-x^2y'=2a^2.\]
Step 2: Rewrite in standard linear form
\[y'-\frac{2}{x}y=-\frac{2a^2}{x^2}.\]
Step 3: Solve the linear equation
With integrating factor $\mu(x)=x^{-2}$: $\dfrac{d}{dx}\!\left(\dfrac{y}{x^2}\right)=-\dfrac{2a^2}{x^4}$, so $\dfrac{y}{x^2}=\dfrac{2a^2}{3x^3}+C$, i.e.
\[y=\frac{2a^2}{3x}+Cx^2.\]
Step 4: Apply the initial condition
At $(a,a)$: $a=\dfrac{2a}{3}+Ca^2\Rightarrow C=\dfrac{1}{3a}$, giving $y=\dfrac{2a^2}{3x}+\dfrac{x^2}{3a}$, i.e. $3axy=2a^3+x^3$.
Result
Final answer:
General solution: \[y=\frac{2a^2}{3x}+Cx^2,\qquad C\in\mathbb{R}\]
Particular solution through $(a,a)$:
\[\boxed{\,3axy=2a^3+x^3.}\]