Question

1. y = c₁e^x + c₂ xe^x ; y” -2y’ + y = 0

2. 2. x2 + y2 = r2 ; y’ = - 𝑥 /y

Answer 1

1. Let us solve the equation $y'' -2y' + y = 0.$ The characteristic equation $k^2-2k+1=0$ is equivalent to $(k-1)^2=0,$ and hence it has the roots $k_1=k_2=1.$ Therefore, its general solution is indeed $y = C_1e^x + C_2 xe^x.$

2. Let us solve the equation $y' = - \frac{𝑥}{y}$ is equivalent to $\frac{dy}{dx} = - \frac{𝑥}{y},$ and hence to $ydy=-xdx.$ It follows that $\int ydy=-\int xdx,$ and consequently $\frac{y^2}{2}=-\frac{x^2}{2}+C.$ Since $C=\frac{y^2}{2}+\frac{x^2}{2}\ge0,$ we conclude that $2C\ge 0,$ and hence $2C=r^2$ for some $r.$ It follows that $\frac{r^2}{2}=\frac{y^2}{2}+\frac{x^2}{2},$ and therefore the general solution of the differential equation is indeed of the form $x^2 + y^2 = r^2.$