(x^2 y^21)^3x^2*y^3=0Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music1302 谁来帮我把这个函数用matlab画出来，挺有趣的。 (x^2 3 1323 请问matlab能绘制隐函数图形吗，怎么Question X^2 Y' 2xy Y^3 = 0 X > 0 Dy/dx = X^2 Y^3/(1 X^3) (2xy^2 2y) (2x^2y 2x) Y' = 0 6y" 5y' Y = 0, Y(0) = 4, Y'(0) = 0 Y" 4y' 4y = 0, Y(0) = 2, Y'(0) = 1 2x^2 Y" 3xy' Y = 0, Y_1 = X^1, X > 0 Y" Y' 3y = 3x^2 3x This problem has been solved! y'=(xy)^2 mit y(0)=0 ich hab schon versucht die homogene Dgl y'2xyy^2=0 durch substituion ( z=y^1) (Bernoullische Differentialgleichung) zu lösen, hat aber nicht funktioniert da ich später e^X^2 integrieren musste Das ist glaub ich auch nicht der richtige ansatz Ich hoffe mal das mir jemand helfen kann mach's doch nicht so schwer substituiere einfach mal ,
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Ich habe bei der Gleichung (x 1) 2 = 0 auf beiden Seiten die Wurzel gezogen √(x 1) 2 = √0 x 1Question X^2 Y' 2xy Y^3 = 0 X > 0 Dy/dx = X^2 Y^3/(1 X^3) (2xy^2 2y) (2x^2y 2x)y' = 0 6y" 5y' Y = 0, Y(0) = 4, Y'(0) = 0 Y" 4y' 4y = 0, Y_1 = X^1, X > 0 Y" Y' 3y = 3x^2 3x Please Do Not Skip My Steps! Here the line y = 0 is the asymptote parallel to Xaxis whereas there is no asymptote parallel to Yaxis For Oblique Asymptotes In the given equation of curve, expression containing the third degree terms is y 3 x 2 y 2xy 2 Thus, φ 3 (m) = m 3 2m 2 m (by taking y = m, x = 1) so that φ' 3 (m) = 3m 2 4m 1 and φ" 3 (m) = 6m 4
0 z xy2 y3g und wir erhalten K(y) = f(x;z) 2R2 0 x 2;X 2 y 2 − 1 = x 2 / 3 y , which can easily be solved for y y = 1 2 ( x 2 / 3 ± x 4 / 3 4 ( 1 − x 2)) Now plot this, taking both branches of the square root into account You might have to numerically solve the equation x 4 / 3 4 ( 1 − x 2) = 0 in order to get the exact x interval Share answered Dec 22 '12 at 1731 ChristianThere are 2 second degree terms so we are sure it is not parabola x^2y^22x3=0 x^22xy^2=3 add 1 to both sides of the equation x^22x1y^2=31 (x^22x1)y^2=4 (x1)^2(y0)^2=2^2 it takes the form (xh)^2(yk)^2=r^2 with center at (1, 0) with radius r=2 See the graph of x^2y^22x3=0 graph{(x1)^2(y0)^2=2^210,10,5,5} God bless I hope theCompute answers using Wolfram's
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Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music WolframAlpha brings expertlevel knowledge andDiesmal beide nach x auflösen und gleichsetzen I y 2x = 16 y 2x = 16 y 2 I' x = 8 0,5y II 3y x 8 = 0 x II' 3y 8 = x I'=II' 8 0,5y = 3y 8 0,5y 8 0 = 3,5y y = 0 In II' x = 8 0 = 8 Probe I y 2x = 16 0 2·8 = 16 OK II 3y x 8 = 0 0 8 8 = 0 OK S 19, Nr 2 a) 2x 5y = 9 y = 3x 12 rechte Seite von II in I für y einsetzen 2x 5y = 9 2x 5 Ex 32,9 Find x and y, if 2 8(1&3@0&𝑥) 8(𝑦&0@1&2) = 8(5&6@1&8) Given that 2 8(1&3@0&𝑥) 8(𝑦&0@1&2) = 8(5&6@1&8) 8(1×2&3×2@0×2&𝑥
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0 z y2x y3g f ur y;2 Die Menge K(y) ein Trapez mit den Eckpunkten (0;0), (2;0), (2;2y2 y3) und (0;y3) bzw ein Dreieck f ur y= 0 und es folgt daher jK(y)j= 2y3 2((y 32y2) y3) 2 = 2y3 2y2, woraus Oder du lässt \(x^2(y\sqrt3{x^2})^2=1\) plotten Kommentiert von racine_carrée Bitte logge dich ein oder registriere dich , um zu kommentierenThis problem has been solved!
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E1=100(3 10 2)3 9 2 e1= 6 0 Daumen Hi, ein Vorschlag wäre der der Gleichsetzung x = 32y x = 43y Gleichsetzen 32y = 43y 3y3 y=1 Damit in Gleichung 1 x = 32 = 1 Damit sind x=1 und y=1 Grüße Beantwortet von Unknown 139 k 🚀 wow das ging aber schnell ,vielen dank für ihre hilfe )) Kommentiert von Gast Ein anderes Problem?Simple and best practice solution for (3x^2y2xyy^3)dx(x^2y^2)dy=0 equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework If it's not what You are looking for type in the equation solver your own equation and let us solve it Equation SOLVE Solution for (3x^2y2xyy^3
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X 1 = 1 (x 3) 2 = 0 x 3 = 0 x 2 = 3 Besten Gruß Beantwortet 8 Dez 13 von Brucybabe 32 k Danke für deine Mühe erstmal )Warum wird von (x1) 2 =0 x1?Пример если x = 10, y = 5a (10 5a) 2 = 10 2 2105a (5a) 2 = 100 100a 25a 2 (10 4) 2 = 10 2 2104 4 2 = 100 80 16 = 36 Конечно, если мы имеем следующую ситуацию 25 a 4a 2 = 5 2 225 (2a) 2 = (5 2a) 2 (x y) 3 = x 3 3x 2 y 3xy 2 y 3 (x y) 3 = x 3 3x 2 y 3xy 2 y 3 Пример (1 a 2) 3 = 1 3 3Solve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and more
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Kommentiert 8 Dez 13 von Gast aaaah habs von alleine verstanden is ok Kommentiert 8 Dez 13 von Gast Gern geschehen!X4y−14 =0,5xy−13= 0 Given the unique solution of equation is x = 2 and y = 3 Put the value of x and y in each of the given equations to check if these values satisfy the equations (A) xy = 1 ,We get 23 = 5 = 1 2x3y =5 2×23×3 = 49 = 13 = 5 (B) 2x5y = 11 2×25×3 ich soll folgendes Anfangswertproblem kösen y' = ( (2xy)/ (1x^2 ) ) , y(0) = 3 dy/ dx = ( (2xy) 3= C y= 3* Ix^2 1I stimmt das?
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I Need The Answers Of The Highlighted Problems!!!3,(0,0)(x,y) = y xy −y 2 −xy2 1 3 y 3 1 Aufgabe 2 a) Es gilt gradf(x,y) = (y 1,x − 2) = (0!,0) genau dann, wenn (x,y) = (2,−1) ist Somit ist (2,−1) der einzige kritische Punkt von f Wegen detH f(2,−1) = det 0 1 1 0 = −1 < 0 ist die HesseMatrix H f(2,−1) indefinit, so dass f in (2,−1) einen Sattelpunkt besitzt b) Der Gradient von f lautet gradf(x,y) = (6x2 − 3yOSMBOY OSMBOY Math Secondary School answered X/22y/3=1 and xy/3=3 solve by elimination method 2 See answers Advertisement Advertisement KrishnaPolavarapu KrishnaPolavarapu Here's your answer Hope it was helpful EthicalElite EthicalElite x/22y/3=1 and xy/3=3 x/22y/3=1 (1) xy/3=3 (2
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