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# Solving 2xyy' = x^2 + y^2

Let's solve the following Homogeneous first-order ordinary differential equation $$2xyy\prime = x^2 + y^2$$. The form of this equation is pretty obvious to make this decision. The power of $$2xy$$ is 2, the power of $$x^2 + y^2$$ is two. So this is a classical homogeneous first-order ODE. And therefore, let's follow the standard procedure:

$2xyy\prime = x^2 + y^2 \\ \frac{\mathop{dy}}{\mathop{dx}} = \frac{x^2 + y^2}{2xy} = \frac{x^2}{2xy} + \frac{y^2}{2xy} = \frac{1}{2\frac{y}{x}} + 2\frac{y}{x} = Q(\frac{y}{x}) \\ y = v\times x \\ \frac{\mathop{dy}}{\mathop{dx}} = x\frac{\mathop{dv}}{\mathop{dx}} + v \\ 2xvx(x\frac{\mathop{dv}}{\mathop{dx}} + v) = x^2 + v^2x^2 \\ 2v(x\frac{\mathop{dv}}{\mathop{dx}} + v) = 1 + v^2 \\ x\frac{\mathop{dv}}{\mathop{dx}} + v = \frac{1 + v^2}{2v} \\ x\frac{\mathop{dv}}{\mathop{dx}} = \frac{1 - v^2}{2v} \\ \frac{2v\mathop{dv}}{1 - v^2} = \frac{\mathop{dx}}{x} \\ \int\frac{2v\mathop{dv}}{1 - v^2} = \int\frac{\mathop{dx}}{x} \\ \int\frac{\mathop{dx}}{x} = \ln{\left|x\right|} + C_1 \\ \int\frac{2v\mathop{dv}}{1 - v^2} = \cdots \\ 1 - v^2 = w \Rightarrow \mathop{dw} = - 2v\mathop{dv} \Rightarrow \mathop{dv} = - \frac{\mathop{dw}}{2v} \\ \int\frac{2v\mathop{dv}}{1 - v^2} = -\int\frac{2v\mathop{dw}}{w2v} = -\int\frac{\mathop{dw}}{w} = \ln{\left|w\right|} + C_2 = -\ln{\left|1 - v^2\right|} + C_2 \\ -\ln{\left|1 - v^2\right|} = \ln{\left|x\right|} + C; C \in \mathbb{R} \\ \ln{\left|x\right|} + \ln{\left|1 - v^2\right|} = C\\ \ln{\left|x - xv^2\right|} = C\\ e^{\ln{\left|x - xv^2\right|}} = e^C = K \Rightarrow K > 0 \\ \left|x - xv^2\right| = K x - xv^2 = \pm K x - xv^2 = A; A \in \mathbb{R} \\ v = \frac{y}{x} \Rightarrow x - x\frac{y^2}{x^2} = A \\ x - \frac{y^2}{x} = A \\ y^2 = x^2 - Ax \Rightarrow y^2 = x^2 + Ax \\ \boxed{y = \pm\sqrt{x^2 + Ax}}$

I'm hoping that all these steps are self-explanatory, but if you feel some confusion, feel free to watch the video with further comments on how to solve $$2xyy\prime = x^2 + y^2$$.

Published: 2023-05-14 21:57:15

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