How do you solve $\frac{dy}{dx}=2-x$ and find the curve through $(1,0)$?

Expert intro: Unlike a separable exponential equation, this one is even simpler: the derivative depends only on $x$, so you can integrate directly and then use the given point.

Detailed walkthrough

1) Integrate the differential equation

$\frac{dy}{dx}=2-x$

Integrate both sides with respect to $x$:

$y=\int (2-x)\,dx$

$y=2x-\frac{x^2}{2}+C$

So the general solution is

$\boxed{y=2x-\frac{x^2}{2}+C}$

2) Use the point $(1,0)$

Substitute $x=1$ and $y=0$:

$0=2(1)-\frac{1^2}{2}+C$

$0=2-\frac12+C$

$C=-\frac32$

So the particular solution is

$\boxed{y=2x-\frac{x^2}{2}-\frac32}$

3) How the sketch should look

The graph is a downward-opening parabola. Its slope is positive when $x<2$, zero at $x=2$, and negative when $x>2$. On the slope field, the line segments flatten as you approach $x=2$ and tilt downward after that.

💡 Pitfall guide

Don’t treat this like a separable equation. There is no $y$ on the right-hand side, so you just integrate with respect to $x$. Another easy error is dropping the constant after integration, which makes the particular solution impossible to fit through $(1,0)$.

🔄 Real-world variant

If the initial point were $(0,0)$, then the constant would be $0$ and the solution would be $y=2x-\frac{x^2}{2}$. If the derivative were $\frac{dy}{dx}=a-x$, the antiderivative would still be a parabola; only the linear coefficient would change.

🔍 Related terms

antiderivative, initial value problem, slope field