Solve the following initial value problem. y' + 2x(y+1) = 0, y(0) = 2

Contact me if you need help with homework, assignments, or Exams for differential equations.
You can see our Testimonials or Vouches from here of the previous works I have done.

Step-by-Step-Solution: We can solve the given differential equation using the variable separable method.
So, after separating the variables

$\frac{1}{y+1}dy = -2xdx$

Now, integrating both sides.

$\int \frac{1}{y+1}dy = \int -2xdx$

$ln(y+1) = -x^2 + c$

We will use the initial condition $y(0) = 2$ to find the value of $c$

$ln(3) = 0 + c$

then $c = ln(3)$

So, the particular solution of the differential equation is

$ln(y+1) = -x^2 + ln(3)$

after simplifying

$ln(\frac{y+1}{3}) = -x^2$

$y = (3e^{-x^2} - 1)$

If this solution helps, then please share this with your friends.
Please subscribe to my Youtube channel for video solutions of similar questions.
Keep Smiling :-)