Question

practicing my first separable DE

Answer 1

\[\large \frac{dy}{dx}=\frac{x^2y-4y}{x+2}\]

Answer 2

\[\large \frac{dy}{dx}=\frac{x^2y-4y}{x+2}\]\[\large \frac{1}{y}\frac{dy}{dx}=\frac{x^2-4}{x+2}\]\[\large \frac{1}{y}\frac{dy}{dx}=x-2\]\[\large\color{red}{\int\limits} \frac{1}{y}\frac{dy}{dx}\color{red}{dx}=\large\color{red}{\int\limits} (x-2)\color{red}{dx}\]\[\large\color{rd}{\int\limits} \frac{1}{y}\color{rd}{dy}=\large\color{rd}{\int\limits} (x-2)\color{rd}{dx}\]\[\ln|y|=\frac{1}{2}x^2-2x+C_1\]\[\Large y=e^{\frac{1}{2}x^2-2x+C_1}\]\[\Large y=e^{C_1}e^{\frac{1}{2}x^2-2x}\]\[\Large y={C_2}~e^{\frac{1}{2}x^2-2x}\]

Answer 3

@tkhunny @rational can you guys check

Answer 4

?

Answer 5

Two things:

1) Your solution is not value for x = -2 or y = 0. You should have stated that.

2) Does anyone actually care if your arbitrary constant changes? It's still just an arbitrary constant. To be fair, some folks do care.

Answer 6

I saw a video that Professor Leonard (if you heard of him) recorded on youtube, and I saw that they had to denote their constants, even though I do agree they are just C 's.

and tnx for the first correction as well

Answer 7

I guess it isn't bad to say C_1 and C_2......
but apparently the process is right; that is the most important for me.

Answer 8

tnx