ordinary differential equations - How to solve $y' + xy^2 = 2xy ...
2016年11月4日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for …
How do I prove $2xy ≤ x^2+y^2$? - Mathematics Stack Exchange
2018年10月7日 · $$2xy ≤ x^2 + y^2$$ I just can end up in $(xy)^{1/2} ≤ (x+y)/2$. And even though I know that this is ...
How do I prove that $x^2 + y^2 = (x + y) ^2 – 2xy$ geometrically ...
2015年5月9日 · The answers offered so far just introduce $-2xy$ without explaining where it comes from (except as the inverse of $+2xy$). Remark: I am perfectly aware that $$(a + b)^2= …
ordinary differential equations - Solution of $ (x^2 + y^2)\ dx -2xy ...
2015年6月29日 · $\frac{dy}{dx}=\frac{x^2+y^2}{2xy} $..(i) This is a homogeneous differential equation because it has homogeneous functions of same degree 2. homogeneous functions …
Solution of the differential equation $y'' + 2xy' + \left( 1 + x^2 ...
2017年10月29日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for …
Solve the differential equation $y'-xy^2 = 2xy$
2015年2月16日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for …
How to prove $2x^2 + 2y^2 + 2z^2 -2xy -2yz \\geq 0$
2019年12月13日 · Find max of $(3x^2-2xy+3y^2)(3y^2-2yz+3z^2)(3z^2-2xz+3x^2)$ Hot Network Questions Is there still an active cryptographic standard in some developing country that …
Find the value of $2xy$ - Mathematics Stack Exchange
2015年7月18日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for …
quadratic forms - show that the function $z = 2x^2 + y^2 +2xy
Now the polynomial becomes $$ 2X^2+2XY+Y^2-3 $$ and now it is clear that its values are $\ge-3$, because $2X^2+2XY+Y^2$ is positive definite (the conic is an ellipse), having negative …
algebra precalculus - How to Factorise $x^2 + y^2 + z^2 - 2xy
2021年7月17日 · In this case the factorisation becomes, $$ x^2 + y^2 + z^2 - 2xy - 2yz - 2zx = (ix + jy + kz)^2. $$ I didn't know if there was some algebra which obeyed these properties. …