Rewrite the given equation as (y*dx x*dy) * (x^2 y^2) (y*dx x*dy) = 0 Eq(1) Divide throughout by (x^2 y^2) (y*dx x*dy) (y*dx x*dy) / (x^2 y^23/1/12 · (dy/dx) √(1y²) / √(1x²) = 0 ∴∫ dy/√(1y²) ∫ dx/√(1x²) = ∫ 0 dx ∴ ( sinֿ¹ y ) ( sinֿ¹ x ) = sinֿ¹ c, saySimple and best practice solution for y(2x^2xy1)dx(xy)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
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X 1 y 2 dx y 1 x 2 dy 000-Wolfram Alpha gives which looks a lot more likely to me to be correct than the other answer provided so far (which includes a pretty elementary error in theHelp is appreciated Edit
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E x root(1y 2)dx y/x dy=0 Share with your friends Share 0 e x 1y 2 d x y x d y = 0 ⇒ y x d y = e x 1y 2追问: 可是答案是(1y^2)/(1y^2)=c 答案错了吗。。 追答: 答案错了 追问: y/(1y²)dy=x/(x²1)dx 这一步怎么倒出来啊 倒了半天没出来。10/11/18 · To ask Unlimited Maths doubts download Doubtnut from https//googl/9WZjCW `x(1y^2)dxy(1x^2) dy=0`
Click here👆to get an answer to your question ️ Solve the differential equation x(1 y^2)dx y(1 x^2)dy = 0But if I expand the bracket $(xy)^2$ before integrating I will get $$\varnothing_1=\int Mdx=\int (xy)^2dx=\int (x^22xyy^2)dx=\frac{x^3}{3}xy^2x^2y$$ Wich will lead to the solution $$\varnothing=\varnothing_1\varnothing_2=\frac{x^3}{3}xy^2x^2yy=Constant$$ What is the wrong step ?Simple and best practice solution for (x^2)dxy(x1)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
10/5/17 · y^2 = ( x^2 1)/2 We have x(1y^2)dxy(1x^2)dy =0 If we rearrange this ODE from differential form into standard form we have y(1x^2)dy/dx =x(1y^2) y/(1y^2)dy/dx =x/(1x^2) This is now a separable ODE, do we can "separate the variables" to give int \\ y/(1y^2)dy = int \\ x/(1x^2) \\ dx We can manipulate the numerators on both sides to get int \\ (2y)/(1y^2)dy = int \\ (2x)/(1Find dy/dx y^2=1/(1x^2) Differentiate both sides of the equation Differentiate the left side of the equation Tap for more steps Differentiate using the chain rule, which states that is where and Tap for more steps To apply the Chain Rule, set as
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