Answer to Question #112783 in Differential Equations for harini

Question

Solvable for y
y²p-2xp+y=0

Answer 1

$$\left.\left(y^2-2x\right)\cdot\frac{dy}{dx}+y=0\right|\times(dx)\\[0.3cm] \underbrace{\left(y\right)}_{M(x,y)}dx+\underbrace{\left(y^2-2x\right)}_{N(x,y)}dy=0\\[0.3cm] \frac{\partial M(x,y)}{\partial y}=1\neq-2=\frac{\partial N(x,y)}{\partial x}$$

Conclusion,

$$\boxed{ydx+\left(y^2-2x\right)dy=0-\text{not exact}}$$

This means that we must find the form of the integrating factor for the equation to be exact.

Since,

$$\varphi(y)=\frac{\displaystyle\frac{\partial M(x,y)}{\partial y}-\displaystyle\frac{\partial N(x,y)}{\partial x}}{-M}=\frac{1-(-2)}{-y}=-\frac{3}{y}$$

The function $\varphi(y)$ is only for the variable $y$ , then the integrating factor has the form

$$\mu(y)=e^{\int\varphi(y)dy}=e^{-\int3/ydy}=e^{-3\ln y}=e^{\ln y^{-3}}=y^{-3}\\[0.3cm] \boxed{\mu(y)=\frac{1}{y^3}}$$

Then,

$$\left(\frac{1}{y^3}\right)\times\left|\left(y^2-2x\right)dy+ydx=0\right.\\[0.3cm] \underbrace{\left(\frac{1}{y^2}\right)}_{F'_x(x,y)}dx+\underbrace{\left(\frac{1}{y}-\frac{2x}{y^3}\right)}_{F'_y(x,y)}dy=0\\[0.3cm] \frac{\partial F}{\partial x}=\frac{1}{y^2}\longrightarrow F(x,y)=\int\frac{dx}{y^2}+g(y)\\[0.3cm] F(x,y)=\frac{x}{y^2}+g(y)\longrightarrow F'_y(x,y)=g'(y)-\frac{2x}{y^3}\\[0.3cm] g'(y)-\frac{2x}{y^3}=\frac{1}{y}-\frac{2x}{y^3}\longrightarrow\frac{dg}{dy}=\frac{1}{y}\longrightarrow\\[0.3cm] \boxed{g(y)=\ln|y|}$$

Conclusion,

$$F(x,y)=\frac{x}{y^2}+\ln|y|\longrightarrow\left(\frac{1}{y^2}\right)dx+\left(\frac{1}{y}-\frac{2x}{y^3}\right)dy=0\\[0.3cm] d\left(\frac{x}{y^2}+\ln|y|\right)=0\longrightarrow\boxed{\frac{x}{y^2}+\ln|y|=Const}$$

This equation defines a function $y(x)$ which is a solution to the original equation