Solve the following differential equation (x^2 y^2) dx – 2xy dy = 0 asked in Differential Equations by Amayra ( 314k points) differential equationsX 2 y xy 2 and g (x, y) = e −() xy 22, derive expressions for the partial derivatives ∂ ∂ ∂ ∂ f g f x xg and 4 Determine whether or not each of the following linear differential forms is an exact differential a dx xy dy 22xy b dx xy dy 22xy c 2 x 2 y dx x 3 dy d ln ydx x y dy 5 Evaluate each ofSolve the differential equation xydx(1x^2)dy=0 Grouping the terms of the differential equation Group the terms of the differential equation Move the terms of the y variable to the left side, and the terms of the x variable to the right side of the equality Simplify the expression \frac{1}{y}dy Simplify the expression x\frac{1}{1x^2}dx
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Obtain the general solution (x^2+y^2)dx+(xy)dy=0-Combine y 2 d x and x y 2 d to get 2 y 2 d x dx^ {3}2y^ {2}dx=0 d x 3 2 y 2 d x = 0 Combine all terms containing d Combine all terms containing d \left (x^ {3}2y^ {2}x\right)d=0 ( x 3 2 y 2 x) d = 0 The equation is in standard form The equation is in standard formIt is x minus two over access And we could also see clearly that X Y goes through zero when X is plus or minus square to to Um And then so this tells us that we have an oblique acetone at along the line X equals four, Y equals X We also have averted class and told the X equals zero You can take a derivative of the second derivative




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solve (x^2y)dy/dx2xy=0 → (2xy) dx (x^2y) dy=0 asked in CALCULUS by homeworkhelp Mentor derivativesThe equation (x^2 y^2 x )dx xydy =0 can be rewritten as dy/dx y/x = (x1)/y which is a Bernoulli equation The reduction to normal form is obtained taking y = V (x)^1/2 Then , dy/dx = (1/2) (V^1/2)V' and the equation becomes V' 2V/x =(x 2 y 2)dx 2xydy = 0 (x 2 y 2) dx = 2xydy `dy/dx = (x^2 y^2)/(2xy)`(i) The equation is a homogeneous equation Let y= vx, Differentiat ing wrt x, we get, `dy/dx=vx(dv)/dx` `dy/dx=(x^2y^2)/(2xy) " from "(i)` `vx(dv)/dx=(x^2(vx)^2)/(2x(vx))` `vx(dv)/dx=(1v^2)/(2v)` `x(dv)/dx=(1v^2)/(2v)v` `x(dv)/dx=(1v^22v^2)/(2v)` `x(dv)/dx=(1v^2)/(2v)`
2 Find an integrating factor and solve the following differential equation (x^2 y^2 x)dx xydy = 0;See the answer See the answer See the answer done loading Solve (x 2 y 2) dx xydy = 0 Best Answer This is the best answer based on feedback and ratings 100% (1 rating) Previous question Next question Get more help from Chegg Find the general solution of y2dx (x2xyy2)dy = 0 0 votes 128k views asked in Class XII Maths by nikita74 Expert (112k points) Find the general solution of y 2 dx (x 2 xyy 2 )dy = 0 differential equations
Get stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!Welcome to Sarthaks eConnect A unique platform where students can interact with teachers/experts/students to get solutions to their queries Students (upto class 102) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (MainsAdvance) and NEET can ask questions from any subject and get quick answers byGiven The homogeneous differential equation is given as, (x2y2)dxxydy =0 ( x 2 y 2) d x x y d y = 0 To find the solution Rewrite the See full answer below




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Solve (x 2y 2)dx3xydy=0 =v In each of the Exercises 1 to 10, show that the given differential equation is homogeneous and solve each of them In each of the Exercises 1 to 10, show that the given differential equation is homogeneous and solve each of them Solve (x^2y^2x)dxxy dy =0 Get the answers you need, now! Ex 95, 6 Show that the given differential equation is homogeneous and solve each of them 𝑥 𝑑𝑦−𝑦 𝑑𝑥=√(𝑥^2𝑦^2 ) 𝑑𝑥 Step 1 Find 𝑑𝑦/𝑑𝑥 x dy − y dx = √(𝑥^2𝑦^2 ) dx x dy = √(𝑥^2𝑦^2 ) dx y dx x dy = (√(𝑥^2𝑦^2 )𝑦) dx 𝑑𝑦/𝑑𝑥 = (√(𝑥^2 𝑦^2 ) 𝑦)/𝑥 Step 2 Put 𝑑𝑦/𝑑𝑥 =




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913 answers 1431K people helped (x^2 xy y^2)dx xydy =0 divide by x^2 (1y/x y^2/x^2) dx y/x dy = 0 Let y/x = v y=vx (1vv^2) dx v dy = 0 dy = (vdxxdv)Learn how to solve differential equations problems step by step online Solve the differential equation (x^23y^2)dx2xydy=0 We can identify that the differential equation \left(x^23y^2\right)dx2xy\cdot dy=0 is homogeneous, since it is written in the standard form M(x,y)dxN(x,y)dy=0, where M(x,y) and N(x,y) are the partial derivatives of a twovariableMultiply y and y to get y 2 \left (x^ {2}y^ {2}\right)dxxy^ {2}d=0 ( x 2 − y 2) d x − x y 2 d = 0 Use the distributive property to multiply x^ {2}y^ {2} by d Use the distributive property to multiply x 2 − y 2 by d \left (x^ {2}dy^ {2}d\right)xxy^ {2}d=0 ( x 2 d − y 2 d) x − x y 2 d = 0



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y^2 = x^2(2lnx c) We can rewrite this Ordinary Differential Equation in differential form (x^2 y^2) \ dx xy \ dy = 0 A as follows \ \ \ \ dy/dx = (x^2 y^2)/(xy) dy/dx = x/y y/x B Leading to a suggestion of a substitution of the form u = y/x iff y = ux And differentiating wrt x whilst applying the product rule dy/dx = u x(du)/dx Substituting into the Transcript Ex 95, 12 For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition 𝑥2𝑑𝑦 𝑥𝑦 𝑦2 𝑑𝑥=0;𝑦=1 When 𝑥=1 The differential equation can be written 𝑎s 𝑥2𝑑𝑦 = −(xy y2) dx 𝑑𝑦𝑑𝑥 = − 𝑥𝑦 𝑦2 𝑥2 Let F(xThey give you the integrating factor is x2, so multiply the whole equation by said factor to get (1y 2 /x 2)dx (12y/x )dy = 0 Now, check for exactness again, dM/dy = dN/dx (these should be partial derivatives) dM/dy = 2y/x 2 dN/dx = 2y/x 2 Therefore this equation is now exact, and solve it like you would any other exact equation



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Solution for Differential Equations Integrating Factors Solve the Equation (x2y2x)dxxydy=0For the differential equation `(x^2y^2)dx2xy dy=0`, which of the following are true (A) solution is `x^2y^2=cx` (B) `x^2y^2=cx` `x^2y^2=xc` (D) `ySolve the Differential Equation (x^2y^2)dxxydy=0 Find where Tap for more steps Differentiate with respect to Solve for Tap for more steps Move all terms containing variables to the left side of the equation Tap for more steps Subtract from both sides of the equation




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The GS is y^2 = (A x^4)/(2x^2) Or, alternatively y = sqrt(A x^4)/(sqrt(2)x) We have (x^2 y^2) \ dx xy \ dy = 0 Which we can write in standard form as dy/dx = (x^2 y^2)/(xy) 1 Which is a nonseparable First Order Ordinary Differential Equation(X^2y^2x)dxxydy=0 Solve for general solution Mathematics Answer Comment 1 answer aksik 14 1 year ago 5 0 Check if the equation is exact, which happens for ODEs of the form if We have so the ODE is not quite exact, but we can find an integrating factor so that d x x d y 2 − y 2 d x x 2 = 0 d x d ( y 2 x) = 0 x y 2 x = C It seems to me that there is a sign mistake somewhere Share Follow this answer to receive notifications edited at 2301 answered at 2256 user



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Solution for Solve the equation (x² – xy y² )dx – xydy = 0 %3D O (v x) el) = (v x) e(G) = c O (v x) = ce(÷)Solution for (x^2y^25)dx (yxy)dy=0 equation Simplifying (x 2 y 2 5) * dx 1 (y xy) * dy = 0 Reorder the terms (5 x 2 y 2) * dx 1 (y xy) * dy = 0 Reorder the terms for easier multiplication dx (5 x 2 y 2) 1 (y xy) * dy = 0 (5 * dx x 2 * dx y 2 * dx) 1 (y xy) * dy = 0 Reorder the terms (5dx dxy 2 dx 3) 1 (y xy) * dy = 0 (5dx dxy 2 dx 3) 1Dy / dx = f (y) Solve dy / dx = sin^2y Formation Of Differential Equations Form the differential equation of the family of curves represented c (y c)^2 = x^3 , where c is a parameter Find the differential equation that represents the family of all parabolas having




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Solve the following differential equation (x2 y2)dx 2xy dy = 0Solution for (x^2y^2)dx= (xy)dy equation Simplifying (x 2 1y 2) * dx = (xy) * dy Reorder the terms for easier multiplication dx (x 2 1y 2) = (xy) * dy (x 2 * dx 1y 2 * dx) = (xy) * dy Reorder the terms (1dxy 2 dx 3) = (xy) * dy (1dxy 2 dx 3) = (xy) * dy Multiply xy * dy 1dxy 2 dx 3 = dxy 2 Solving 1dxy 2 dx 3 = dxy 2 Solving for variable 'd'General solution of (x 2 y 2 ) dx 2xydy = 0 is y 2 x 2 = cx 2 x 2 y 2 = cx



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Show activity on this post solve the following differential equation (1) ( x 2 − 2 x 2 y 2) d x 2 x y d y = 0 d y d x = 2 x − x 2 − 2 y 2 2 x y dividing (1) throughout by y 2 we have, (2) ( x 2 y 2 2 − 2 x y 2) d x 2 ( x y) d y = 0 ordinarydifferentialequations ShareQuestion Solve (x2 y2 ) dx xydy = 0 This problem has been solved!The solution You have entered src 2 2 d x y (x) x*y (x) x* (y (x))*y (x) = 0 dx x 2 x y ( x) d d x y ( x) x y ( x) − y 2 ( x) = 0 x^2 x*y*y' x*y y^2 = 0 Detail solution Given the equation x 2 x y ( x) d d x y ( x) x y ( x) − y 2 ( x) = 0 Do replacement



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The equation (x^2 y^2 x )dx xydy =0 can be rewritten as dy/dx y/x = (x1)/y which is a Bernoulli equation The reduction to normal form is obtained taking y = V (x)^1/2 Then , dy/dx =Question 2 Find an integrating factor and solve the following differential equation (x^2 y^2 x)dx xydy = 0The solution of the differential equation dy / dx = xy y / xy x is The solution of the differential equation x (dy / dx) 2y = x 2 is The solution of the equation (3 2 √2) x^28 (3 2 √2) 8x^2 = 6 are The Solution Of The Equation Sin X Cos X Power 1 Sin 2x 2 Pi X Pi Is The Solution Of Trigonometric Equation Cos 4 X Sin 4



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Solution of the differential equation `(x^(2)2y^(2))dxxy dy=0,` when y (9)=0 isExample Solve the ODEs (x 2 y 2) dx2 xydy = 0 Here ∂ M ∂ y∂ N ∂ x N = 2 y 2 y2 xy =2 x (7) Then integrating factor is e R2 x dx = 1 x 2 Now multiplying by 1 x 2 (x 2 y 2) dx2 xy dy x 2 = 0 ⇒ d x 2y 2 x = 0 (8) which is exact and its solution is x 2y 2 = cx Ratikanta Behera Ordinary Differential Equations I September 1 Follow 0 Rishabh Mittal, Meritnation Expert added an answer, on 28/9/15 Rishabh Mittal answered this x x 2 y 2 y 2 dx xy dy = 0 x x 2 y 2 y 2 dx = xy dy dy dx = x x 2 y 2 y 2 xy = y 2 x x 2 y 2 xy = y 2 x 2 1 y 2 x 2 y x So , it is a homogeneous differential equation of degree zero Let y = vx dy dx = v x dv dx v




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