Answer to Question #136589 in Differential Equations for Sohan kumar

Question

Solve dy/dx +2xy= 1+x²+y² given y'(x)=x is a solution.

Answer 1

$\displaystyle\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} + 2xy &= 1 + x^2 + y^2 \\ \frac{\mathrm{d}y}{\mathrm{d}x} &=1 + x^2 + y^2 - 2xy\\ &= 1 + (x - y)^2 \end{aligned}\\ \therefore \frac{\mathrm{d}y}{\mathrm{d}x} = 1 + (x - y)^2 \\ \textsf{Making the substitution} \\ u = x - y \\ \frac{\mathrm{d}u}{\mathrm{d}x} = 1 - \frac{\mathrm{d}y}{\mathrm{d}x}\\ \therefore 1 - \frac{\mathrm{d}y}{\mathrm{d}x} = -(x - y)^2 \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} = -u^2\\ \textsf{Seperating the variables, we have;}\\ -\frac{\mathrm{d}u}{u^2} = \mathrm{d}x\\ \textsf{Integrating both sides, we have;}\\ -\displaystyle\int\frac{\mathrm{d}u}{u^2} = \displaystyle\int \mathrm{d}x\\ \frac{1}{u} = x + C\\ u = \frac{1}{x + C}\\ x - y = \frac{1}{x + C}\\ (x - y)(x + C) = 1\\ y = x - \frac{1}{x + C}\\ \therefore y = x - \frac{1}{x + C} \hspace{0.1cm} \textsf{is also a solution to the}\\\textsf{first order linear differential equation}\\ \textsf{Given that}\hspace{0.1cm} y = x\hspace{0.1cm}\textsf{is also a solution}\\\textsf{to the D.E, then}\\ y = \left(x - \frac{1}{x + C}\right)+ x \hspace{0.1cm} \textsf{is a solution to the}\\\textsf{first order linear differential equation}\\ \therefore y = 2x - \frac{1}{x + C} \textsf{is a solution to the D.E}.$