Mathematics - Differential Equation Question with Solution | TestHub

Let $(x,y)$ be a point on the curve. The tangent at $(x,y)$ has slope $dy/dx$.

The equation of the tangent is $Y-y=\frac{dy}{dx}(X-x)$.

The tangent meets the $y$-axis when $X=0$.

So, $Y-y=\frac{dy}{dx}(0-x)\Rightarrow Y=y-x\frac{dy}{dx}$.

Let $P=(x,y)$ and $Q=(0,y-x\frac{dy}{dx})$.

The midpoint $M$ of $PQ$ is $\left(\frac{x+0}{2},\frac{y+(y-x\frac{dy}{dx})}{2}\right)=\left(\frac{x}{2},y-\frac{x}{2}\frac{dy}{dx}\right)$.

Given that $M$ lies on $y=x$, we have $y-\frac{x}{2}\frac{dy}{dx}=\frac{x}{2}$.

This simplifies to $2y-x\frac{dy}{dx}=x$, or $x\frac{dy}{dx}=2y-x$.

So, $\frac{dy}{dx}=\frac{2y-x}{x}=2\frac{y}{x}-1$.

This is a homogeneous differential equation. Let $y=vx$, so $\frac{dy}{dx}=v+x\frac{dv}{dx}$.

Substituting this into the differential equation:

$v+x\frac{dv}{dx}=2v-1$

$x\frac{dv}{dx}=v-1$

$\frac{dv}{v-1}=\frac{dx}{x}$

Integrating both sides:

$\int \frac{dv}{v-1}=\int \frac{dx}{x}$

$\ln |v-1|=\ln |x|+C$

$\ln \left|\frac{y}{x}-1\right|=\ln |x|+C$

$\frac{y}{x}-1=Ax$, where $A=\pm e^C$.

$y-x=A{x}^2$

$y=A{x}^2+x$.

The curve passes through $(1,0)$. Substitute $x=1,y=0$:

$0=A(1{)}^2+1$

$0=A+1\Rightarrow A=-1$.

Therefore, the equation of the curve is $y=-x^2+x$, or $y=x-x^2$.

The final answer is $\boxed{C}$