You don't: your solution is a special case that SyberMath ignored in his solution. When the u and z variables were separated, both sides of the equation were divided by 1+u². SyberMath ignored the possibility that this is zero. (He tends to ignore special solutions, presumably to simplify and shorten the video.) So another, special solution to the differential equation is 1+u² = 0. This is equivalent to u = ±i. Since w = uz, w = y-1, and z = x+4, we get the special solutions y - 1 = ±i (x + 4) Your solution is the plus option of that equation. Yet another solution is the minus option, y = -ix-4i+1.
Such a complicated equation for such a simple derivative.
You can develop the solution a bit: arctan[(y-1)/(x+4)]=(1/2)*ln [(x+4)^2+(y-1)^2] + C, therefore
e^arctan[(y-1)/(x+4)]=k*[(x+4)^2+(y-1)^2]^1/2.
👍👍👍👍😊😊😊😊
I got a much simpler result using exact equations. But the method seems like it should work. What's the difference?
Very nice! And here is ahomework for you syber: y=ix+4i+1 solves this DE,but how can we get this solution from the nice general solution you found?😊💯
You don't: your solution is a special case that SyberMath ignored in his solution.
When the u and z variables were separated, both sides of the equation were divided by 1+u². SyberMath ignored the possibility that this is zero. (He tends to ignore special solutions, presumably to simplify and shorten the video.) So another, special solution to the differential equation is 1+u² = 0. This is equivalent to u = ±i. Since w = uz, w = y-1, and z = x+4, we get the special solutions
y - 1 = ±i (x + 4)
Your solution is the plus option of that equation. Yet another solution is the minus option, y = -ix-4i+1.
Ez. Set y = ax + b and y’ = a. Plug it in and get the ratios. Then solve for a and b. 😁
Good point! See my reply to @yoav613