Question

find differential equation
$\frac{dy}{dx} = \frac{1}{(x+3)^2}$

Explanation:

Step1: Recognize the type of DE

This is a separable differential equation. We can separate variables and integrate both sides. The equation is $\frac{dy}{dx}=\frac{1}{(x + 3)^2}$.

Step2: Separate variables

Multiply both sides by $dx$ to get $dy=\frac{1}{(x + 3)^2}dx$.

Step3: Integrate both sides

Integrate the left side with respect to $y$ and the right side with respect to $x$.
$\int dy=\int\frac{1}{(x + 3)^2}dx$
For the right - hand integral, let $u=x + 3$, then $du=dx$. So $\int\frac{1}{u^2}du=\int u^{-2}du$.
Using the power rule for integration $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$), we have $\int u^{-2}du=\frac{u^{-2 + 1}}{-2+1}+C=\frac{u^{-1}}{-1}+C=-\frac{1}{u}+C$.
Substituting back $u = x + 3$, we get $\int\frac{1}{(x + 3)^2}dx=-\frac{1}{x + 3}+C$.
The left - hand integral $\int dy=y + C_1$.
Combining the two results: $y=-\frac{1}{x + 3}+C$, where $C = C - C_1$ (a new constant of integration).

Answer:

The general solution of the differential equation $\frac{dy}{dx}=\frac{1}{(x + 3)^2}$ is $y=-\frac{1}{x + 3}+C$