How do you solve the differential equation $\frac{dy}{dx}=y$ and pass through $(1,1)$?

Expert intro: This one is separable, and the solution curve is one of the simplest exponential models. The slope field should line up with the exponential shape you get from the algebra.

Detailed walkthrough

1) Solve the differential equation

$\frac{dy}{dx}=y$

Separate variables:

$\frac{1}{y}\,dy=dx$

Integrate both sides:

$\int \frac{1}{y}\,dy=\int dx$

$\ln|y|=x+C$

Exponentiate:

$|y|=e^{x+C}=Ae^x$

So the general solution is

$y=Ce^x$

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

Substitute the point into the general solution:

$1=Ce^1$

$C=e^{-1}$

So the particular solution is

$\boxed{y=e^{x-1}}$

3) What to sketch on the slope field

The slope at each point $(x,y)$ equals $y$, so:

• slopes are flat along $y=0$
• slopes get steeper as $y$ increases
• slopes tilt downward when $y<0$

The solution through $(1,1)$ should rise exponentially and pass exactly through that point.

💡 Pitfall guide

It’s easy to stop at $\ln|y|=x$ and forget the constant of integration. That leads to a missing family of solutions. Another common mistake is writing $y=e^x$ too early before using the point $(1,1)$ to fix the constant.

🔄 Real-world variant

If the initial point were $(0,1)$ instead, the same method gives $y=e^x$. If the equation were $\frac{dy}{dx}=ky$, the solution would become $y=Ce^{kx}$, so the growth rate changes but the process stays the same.

🔍 Related terms

slope field, separable differential equation, exponential function