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  Equation Solvable for p  MathJax Example Solve \(x^2 \left(\frac{dy}{dx}\right)^2 - 2xy \frac{dy}{dx} + 2y^2 - x^2 = 0\). Solution: \[ \begin{align*} & \text{The given equation is} \\ & x^2 \left(\frac{dy}{dx}\right)^2 - 2xy \frac{dy}{dx} + 2y^2 - x^2 = 0 \\ & \text{or} \\ & \left(\frac{dy}{dx}\right)^2 - 2xy \frac{dy}{dx} + 2y^2 - x^2 = 0 \\ & \text{or} \\ & \left(\frac{dy}{dx}\right) = \frac{2xy \pm \sqrt{4x^2 y^2 - 4x^2 (2y^2 - x^2)}}{2x^2} \\ & \text{or} \\ & \left(\frac{dy}{dx}\right) = \frac{y \pm \sqrt{x^2 - y^2}}{x} \\ \end{align*} \] Put \(y = vx\) so that \(\frac{dy}{dx} = v + x \frac{dv}{dx}\). Therefore, (2) becomes \(v + x \frac{dv}{dx} = \frac{vx \pm \sqrt{x^2 - v^2 x^2}}{x}\). Therefore, \(v + x \frac{dv}{dx} = v \pm \sqrt{1 - v^2}\). Therefore, \(x \frac{dv}{dx} = \pm \sqrt{1 - v^2}\). Separating the variables, \(\frac{1}{\sqrt{1 - v^2}} dv = \pm \frac{1}{x} dx\). Ma...

Solution of (x² + y² 2xy)dx +2ydy=0 is a non-exact differential EQUATION  In this page, now you have to see that the Question (x^2 + y^2 + 2x)dx + 2ydy = 0 is the non exact differential equation and here is the solution of this question. In our website you have to see that we are updated higher Maths question and videos on it, that why you can not confuse for understand the question. Here, we find the integrating factor of this equation and also we solve the full question. This is the differential equation equation and we will solve it. SOLUTION: (x^2 +y^2 +2x)dx + 2ydy + 0 Comparing this with Mdx + Ndy = 0 Here, M = x^2 + y^2 + 2x and N= 2y dM/dy = 2y and dN/dx = 0 (dm/dy-dn/dx)/N = (2y-0)/2y = 1 or x^0 or f(x) e^integral f(x)dx = e^integral x^0dx = e^integral dx = e^x Now, Multiplying throughout in above equation e^x(x^2 + y^2 + 2x)dx + e^x2ydy = 0 Which is Exact differential equation integral e^x x^2 dx + integral e^xy^2dx + integral e^x 2xdx = c x^2 e^x - integral 2xe^x dx...