Show That The Given Differential Equation Is Homogeneous And Solve Each Of Them X2 Y2 Dx 2xy Dy 0 Mathematics Shaalaa Com
Correction (after missing a sign) As kobe pointed out, the original DE is $$ (x^2y^2)y'2xy=0, $$ which as equation for a vector field reads $$ (x^2y^2)\,dy2xy\,dx=0\iff Im(\bar z^2\,dz)=0\text{ with } z=xiy $$ From the complex interpretation it is directly visible that this is not integrable, for that it would have to be an expression $Im(f(z)\,dz)$ But since $\barSo MxNy=4x Now, (MxNy)/N=4x/ (2xy)=2/y, so this is just a function of y v (y) That means that we can let u (y)=e integral of vdy u (y)=e 2ln y =1/y 2 So the equation (x 2 /y 2 1)y'2x/y=0 is exact Now, that means that, letting (x 2 /y 2 1)=fy, 2x/y=fx, then f=x 2 /yyc (x) and f=x 2 /yd (y) These two equations match up, so we
X 2 y 2 2xy-4x-4y 4 0
X 2 y 2 2xy-4x-4y 4 0-F ( x, y) = 2 a x 2 x 2 a 2 x 2 = 2 a 1 a 2 This is a constant ≠ 0, so as x → 0, f ( x, a x) does not convege to f ( 0, 0) Thus your function is not continuous at ( 0, 0) Regarding the one variable functions g ( x) = 2 x b x 2 b 2 and h ( x) = 2 a x a 2 x 2, it's easier I'll include more steps than you might want to write when you get more experience, but here we go x^22xyy^2x=2 Differentiate both sides with respect to x d/(dx)(x^22xyy^2x) = d/(dx)(2) (Write this every time) d/(dx)(x^2)d/(dx)(2xy)d/(dx)(y^2)d/(dx)(x) = d/(dx)(2) Remember that y is the name of some function(s) of x that I haven't found expressions
X2 Y2 Ey2 X2 Solution Fx
Plot 3x^22xyy^2=1 Natural Language; Stack Exchange network consists of 178 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers Visit Stack ExchangeShare Copy Copied to clipboard x^{2}2yxy^{2}=0 All equations of the form ax^{2}bxc=0 can be solved using the quadratic formula \frac{b±\sqrt{b^{2}4ac}}{2a} The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction x=\frac{2y±\sqrt{\left(2y\right)^{2}4y^{2}}}{2}
Solution for X^2y^2=2xy equation Simplifying X 2 1y 2 = 2xy Solving X 2 1y 2 = 2xy Solving for variable 'X' Move all terms containing X to the left, all other terms to the right Add 'y 2 ' to each side of the equation The factors are (xy)(xy) or (xy)^2 We need to factor the trinomial x^22xyy^2 The factors of x^2 = (x)(x) The factors of y^2 = (y)(y) Since the second sign is positive we are adding the factors meaning the signs of the factors need to be the same Since the first sign is negative both signs must be negative`2xy (dy)/(dx) = x^(2) 3y^(2)`
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