X2+y2 2x+6y+100 Find X2+y2

12 f(x,y) = x2 2y2,x2 y2 ≤ 4 Note that we are dealing with an inequality for the constraint We can consider any point in oronthe boundary of a circle with radius 2.

X2+y2 2x+6y+100 find x2+y2. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction x^ {2}2xy^ {2}6y=0 x 2 2 x y 2 − 6 y = 0 This equation is in standard form ax^ {2}bxc=0 Substitute 1 for a, 2 for b, and y\left (6y\right) for c in the quadratic formula, \frac {. Check to see if f y is equal to N f y = −x 2g0(y) = 6y −x2 3 so that g0(y) = 6y2 3 That gives g(y) = 2y3 3y Put this back in to get the full solution, f(x,y) = c x3 −x2y 2x2y3 3y = C 3 Problem 4 (2xy2 2y)(2x2y 2x)dy dx = 0 Check for “exactness”. X212xy36y2 Final result (x 6y)2 Step by step solution Step 1 Equation at the end of step 1 ((x2) 12xy) (22•32y2) Step 2 Trying to factor a multi variable polynomial 21.

 4 Hint Assume that a polynomial of the form y ( x) = x n a n − 1 x n − 1 ⋯ a 0 satisfies the ODE, and pug it in the ODE Then you shall find that n = 2 Next, you shall find that y = x 2 − 1 / 3 Observe that if y is a solution, then so is c y, for every c ∈ R Satisfaction of the condition y ( 1) = 2, implies that the sought. Apply a linear substitution v' = t sin (2v t) 1/2, v (0) = pi/2 Solve a firstorder homogeneous equation through a substitution solve x y' = y* (log (x) log (y)) Make general substitutions solve 2 t^3 y' (t) = 1 sqrt (1 4 t^2 y (t)) y' (x) = (1x cos (y (x))) cot (y (x)) More examples. X = 6x2 6xy 24x;f y = 3x2 6y To find the critical points, we solve f x = 0 =)x2 xy 4x= 0 =)x(x y 4) = 0 =)x= 0 or x y 4 = 0 f y = 0 =)x2 2y= 0 When x= 0 we find y= 0 from the second equation In the second case, we solve the system below by substitution x y 4 = 0;x2 2y= 0 =)x2 2x 8 = 0 =)x= 2 or x= 4 =)y= 2 or y= 8 The three critical.

Share Multiply \frac {y^ {2}2xyx^ {2}} {x^ {2}y^ {2}} times \frac {2x} {xy} by multiplying numerator times numerator and denominator times denominator Cancel out x in both numerator and denominator Factor the expressions that are not already factored Cancel out xy in both numerator and denominator.  The standard eqn of a circle with centre (a,b) and radius r is (x − a)2 (y −b)2 = r2 for x2 y2 −2x 4y − 4 = 0 we need to complete the square x2 y2 −2x 4y − 4 = 0 x2 −2x y2 4y − 4 = 0 (x2 − 2x 12) (y2 4y 22) −12 − 22 −4 = 0 (x −1)2.  1(x2 2x 1 −1) 1(y2 6y 9 − 9) = 6 1(x2 2x 1) −1 1(y2 6y 9) − 9 = 6 (x 1)2 (y 3)2 − 10 = 6 (x 1)2 (y 3)2 = 16 In the form (x −a)2 (y − b)2 = r, the radius is given by √r and the centre is at (a,b) Then, the centre is at ( −1, − 3) and the radius measures 4 units Hopefully this helps!.

Verified by Toppr Given the equation of the circle is 2x 22y 2−x=0 or, x 2y 2− 2x =0 or, x 2−2x 41 4 21. Steps for Solving Linear Equation 2 x 6 y = 12 2 x 6 y = 1 2 Subtract 6y from both sides Subtract 6 y from both sides 2x=126y 2 x = 1 2 − 6 y Divide both sides by. (b) x2 y00 −2xy0 2y= 0 This equation is linear because it can be written in the form (1) as y00 − 2 x y0 2 x2 y=0 where p(x)=2/x, q(x)=2/x2,f(x) = 0 are continuous on any interval that does not contain x= 0 For example, we could take I=(0,∞) 63.

Y 00 y0 y = x2 2x Sol The equation is equivalent to y00 4y0 4y = 4x2 8x The characteristic equation m2 4m 4 = (m 2)2 = 0 has a root m = 2 with multiplicity 2 The complementary solution is y c = C 1e 2x C 2xe 2x From the quadratic function g(x) = 4x2 8x we assume a quadratic function y. Example 126 Find the solution y y x of y 2y 5y 0, with the initial values y 0 2 y 0 1 The auxiliary equation r2 2r 5 0 has the solutions r 1 2i Thus the general solution is y e x Acos 2x Bsin 2x To solve for A and B using the initial values we must first differentiate y (1227) y e x Acos 2x Bsin 2x e x 2Asin 2x 2os 2x. 1 Cex2 2 Find the general solution of y0 = 4y 2ex √ y Answer y = Ce2x − ex 2 3 Find the general solution of xy0 y = y2 ln x Answer y = 1 1 ln x Cx Part VII Homogeneous equations;.

Question (b) x 2y – 2z 3w = 4 2x 4y – 3z 7w = 10 x – 2y 2 – 4w = 6 z = (3x 6y – 5z 10w = 14 = = x 2y – 2z 3w = 0 2x 4y – 3z 7w = 0 x – 2y 2 – 4w = 0 (3x 6y – 5z 10w = 0 = In #1(b) you solved a nonhomogeneous system with the same coefficient matrix Compare your answer for the above system. Click here👆to get an answer to your question ️ Show that the circles x^2 y^2 8x 2y 8 = 0 and x^2 y^2 2x 6y 6 = 0 touch each other and find the point of contact. So, g(x;0) = 2x2 with 0 x 2 attains a maximum value of 0 when x= 0 and a minimum value of 8 when x= 2 On the boundary QR, we have y= 2x 4 with 0 x 2 And, g(x;.

• 4y= 2yλ (10) • 2x y2= 1 (11) •From (9) we have x=0 or λ=1 If x=0, then (11) gives y=±1 If λ=1, then y=0 from (10), so then (11) gives x=±1 Therefore f has possible extreme values at the points (0,1), (0,1), (1,0), (1,0) Evaluating f at these four points, we find that.  "Now for x 2 y 22x4y=0 you can solve in another way You can add both sides with 5 and you 'll get x 2 y 2 2x 4y 5 = 5 and by factorizing you will end up in the form (x1) 2 (y2) 2 = 5 but r 2 =5 so r = √5, so the coordinates are K(1,2) and r=√5Generally all you need is to factorize it to the form of (X Xo) 2 ( Y Yo ) 2 = r 2 to find both radius and coordinates. X^2y^2–2x6y10 = 0 Or, (x^2–2x1) (y^26y9) = 0 Or, (x1)^2 (y3)^2 = 0 So, (x1)^2 = 0 ==> x = 1 And, (y3)^2 = 0 ==> y = 3 Then, x^2y^2 = (1)^2 (3)^2 = 19 = 10 (answer).

Graph x^2y^22x=0 Find the standard form of the hyperbola Tap for more steps Complete the square for Tap for more steps Use the form , to find the values of , , and Consider the vertex form of a parabola Substitute the values of and into the formula Simplify the right side. ODEs Find all solutions to the ODE y" 6y' 9y = 0 The characteristic equation for this ODE has a double root We use variation of constants to obtai. Solution for 2X6Y=2 equation Simplifying 2X 6Y = 2 Solving 2X 6Y = 2 Solving for variable 'X' Move all terms containing X to the left, all other terms to the right Add '6Y' to each side of the equation 2X 6Y 6Y = 2 6Y Combine like terms 6Y 6Y = 0 2X 0 = 2 6Y 2X = 2 6Y Divide each side by '2' X = 1 3Y Simplifying X.

Simple and best practice solution for X^2y^22x6y10=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. Free system of equations calculator solve system of equations stepbystep. find general solution 1 Find the general solution of y0 = y2 xy y2 Answer x− yln y = Cy 2 Find the general solution of xyy0 = x2ey/x y2 Answer.

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. Get stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!. Worked Solutions 95 Plugging in a convenient value for x , say x = π/4 so that 2x = π/2, we have W π 4 = 1 cos π 2 sin π 2 0 −2sin π 2 2cos π 2 0 −4cos π 2 −4sin.

 So to find our minimum we can use differentiation A' = 10x 40 The minimum will occur when the derivative equals 0 0 = 10x 40 0 = 10 (x 4) x= 4 Therefore, the minimum value will occur when y = 2 and x = 4 This means the minimum value is 2^2. Solve x^2 y′′ − 4xy′ 6y = 2x^4 x^2 , x > 0 6 Solve ( e^(y^2) − 5ysin( xy) x^(−1/2) )dx (2xyey2 − 5x sin(xy) − y^(−1/2))dy = 0 7 Find a power series solution about the point 0 to the equation y′′ − 3x^2 y = 0 8 A mass weighing 16 pounds is attached to a 5footlong spring At equilibrium the spring. MATH 3321 Sample Questions for Exam 2 Second Order Nonhomogeneous Differential Equations Section 34, 35 1 z1(x) = 2x3 xln x, z2(x) = xln x − x3 are solutions of a second order, linear nonhomo geneous equation Ly = f(x) y1(x) = x−2 is a solution of the corresponding reduced equation Ly = 0.

To find the xintercept (s), substitute in 0 0 for y y and solve for x x Solve the equation Tap for more steps Rewrite the equation as 2 x − 10 = 0 2 x 10 = 0 Add 10 10 to both sides of the equation Divide each term by 2 2 and simplify Tap for more steps Divide each term in. 2x 4) = x2( 22x4) 32x = 22x 2x2, for 0 x 2 The critical numbers of 2x3 2x for 0 x 2 are x= 0 and x= 2 3 So, ghas a minimum of 0 at x= 0 and a maximum of 8 27 at x= 2 3;y= 8 3 on. Example 5 (f(x,y) = 2x3 − 6xy y2 4y) Find all critical points of f(x,y) = 2x3 − 6xy y2 4y Solution The first order partial derivatives are fx = 6x2 − 6y fy = −6x 2y 4 So the critical points are the solutions of 6x2 − 6y = 0 −6x 2y 4 = 0 We can rewrite the first equation as y = x2, which expresses y as a function of x We can.

 (i) Find the equation of a circle concentric with the circle `x^(2)y^(2)8x6y10=0` and passes through the point (2,3) (ii) Find the equation of c asked in Mathematics by DivyanshuKumar ( 640k points).  Calculus questions and answers Consider the equation below x^2 y^2 − 2x − 6y − z 10 = 0 Reduce the equation to one of the standard forms Classify the surface (as one of the below) ellipsoid elliptic paraboloid hyperbolic paraboloid cone hyperboloid of one sheet hyperboloid of two sheets Sketch the surface. Answer (1 of 6) We find the intersection of the graphs Subs y = 2x k into the circle equation If the line is a tangent there will be only 1 solution, the discriminant must be zero This sounds contradictory but it means that if k = 3√10 10 then the line is a tangent to the circle at poin.

X^2 y^2 10x 4y 13 = 0 This equations is not in standard form Standard form of the equation of a circle is (xh)^2 (yk)^2 = r^2 where (h,k) are the coordinates of the center of the circle and r is the radius of the circle You need to convert your equation into.  To find the x or y intercepts, simply plus in one of them with 0 xintercept 2x 6 (0) 10 = 0 10 2x = 10 /2 x = 5 yintercept 2 (0) 6y 10 = 0 10 6y = 10. Answer (1 of 2) x(x^4 2x^2y^2 y^4)(ydx xdy) y^6dy = 0 (x^5 2x^3y^2 xy^4)(ydx xdy) y^6dy = 0 x^5ydx 2x^3y^3dx xy^5dx x^6dy 2x^4y^2dy x^2y^4dy y^6dy = 0 Regroup (x^5y 2x^3y^3 xy^5)dx (x^6 2x^4y^2 x^2y^4 y^6)dy = 0 (x^6 2x^4y^2 x^2y^4.

Graph x^22xy^26y6=0 Add to both sides of the equation Complete the square for Tap for more steps Use the form , to find the values of , , and to those of the standard form The variable represents the radius of the circle, represents the xoffset from the origin, and represents the yoffset from origin The center of the circle. Find the Center and Radius x^2y^22x=0 Complete the square for Tap for more steps Use the form , to find the values of , , and represents the xoffset from the origin, and represents the yoffset from origin The center of the circle is found at Center These values represent the important values for graphing and analyzing a circle.  By shifting the origin to suitable point axes remaining parallel, reduce the equation 2x 2 – y 2 – 4x 4y – 3 = 0 reduces to form X 2 / a 2 – Y 2 / b 2 = 1 (a > 0 and b > 0) Find the coordinates of point at which the origin is shifted.

 to obtain this form complete the square on both the x and y terms x2 −2x y2 6y = 26 x2 2( − 1)x1 y2 2(3)y9 = 2619 (x −1)2 (y 3)2 = 36 ← in standard form with centre = (1, − 3) radius = √36 = 6 graph { ( (x1)^2 (y3)^236)=0 , , 10, 10} Answer link.  Explanation #x^2 6x y^2 4y = 12 Then take 1/2 of the 'b' term for both quadratic expressions, square those values and add them to both sides #x^2 6x 9 y^2 4y 4 = 12 9 4 (x 3)^2 (y 2)^2 = 25 Circle centered. Move −1 1 to the right side of the equation by adding 1 1 to both sides (x1)2 y2 −6y = −91 ( x 1) 2 y 2 6 y = 9 1 Complete the square for y2 −6y y 2 6 y Tap for more steps Use the form a x 2 b x c a x 2 b x c, to find the values of a a, b b, and c c.