Question

Someone please answer this xy²+(x-1)y-1

Answer 1

math]\begin{equation}\begin{split}(x+y+1)^2\dfrac{dy}{dx}=1\end{split}\end{equation}\tag*{}[/math]

To solve this differential equation, the best trick is to substitute [math]x+y+1=t[/math]

Let’s now take the derivative of both sides…

[math]dx+dy=dt[/math]

[math]\implies 1+\dfrac{dy}{dx}=\dfrac{dt}{dx}[/math]

[math]\implies \dfrac{dy}{dx}=\dfrac{dt}{dx}-1[/math]

So now, our differential equation becomes…

[math]\begin{equation}\begin{split}(x+y+1)^2\dfrac{dy}{dx}=1\\t^2\left(\dfrac{dt}{dx}-1\right)=1\\\left(\dfrac{dt}{dx}-1\right)=\dfrac{1}{t^2}\\\dfrac{dt}{dx}=\dfrac{1}{t^2}+1\\\dfrac{dt}{dx}=\dfrac{1+t^2}{t^2}\\\implies \dfrac{t^2dt}{t^2+1}=dx\end{split}\end{equation}\tag*{}[/math]

Let’s now integrate both sides…

[math]\begin{equation}\begin{split}\int\dfrac{t^2}{t^2+1}\,dt=\int\,dx\\\int\dfrac{(t^2+1)-1}{t^2+1}\,dt=\int\,dx\\\int\,dt-\int\dfrac{1}{t^2+1}\,dt=\int\,dx\\t-\tan^{-1}t=x+C\\ \not{x}+y+1-\tan^{-1}(x+y+1)=\not{x}+C\\\implies \boxed {y+1-\tan^{-1}(x+y+1)=C}\end{split}\end{equation}\tag*{}[/math]

Answer 2

xy²+(x-1)y-1

xy2 + xy - x - y + 1