Question

The solution of the differential equation
dy/dx=y with y(0)=0
exists,but is not unique.
true or false with complete explanation

Answer 1

$\displaystyle\textsf{Given that}\hspace{0.1cm} F(x, y) =\frac{\mathrm{d}y}{\mathrm{d}x} = y, \\F(x, y)\hspace{0.1cm}\textsf{is a continuous function.}\\ \displaystyle\frac{\partial F(x, y)}{\partial y} = 1. \\ \displaystyle\textsf{Since}\hspace{0.1cm}F(x,y) \hspace{0.1cm}\textsf{and}\hspace{0.1cm} \frac{\partial F(x, y)}{\partial y}\hspace{0.1cm} \textsf{are continuous}\\\textsf{on all points}\hspace{0.1cm} (x,y), \hspace{0.1cm}\textsf{so by uniqueness}\\\textsf{theorem we can conclude that a}\\\textsf{solution exists in some open interval}\\\textsf{centered at}\hspace{0.1cm}0,\hspace{0.1cm} \textsf{and is unique in some}\\\textsf{interval centered at}\hspace{0.1cm} 0. \hspace{0.1cm}\textsf{By separating}\\\textsf{variables and integrating, we derive}\\\textsf{a solution to this differential equation.}\\ \frac{\mathrm{d}y}{y} = \mathrm{d}x\\ \int\frac{\mathrm{d}y}{y} = \int \mathrm{d}x\\ \ln{y} = x + C\\ y = e^C \cdot e^x\\ y = Ae^x\hspace{0.3cm} (A = e^C)\\ y = Ae^x\\ \textsf{at}\hspace{0.1cm}x = 0, y = A\\ (0, A) \hspace{0.1cm}\textsf{is a point on}\hspace{0.1cm}y, \hspace{0.1cm}\textsf{where}\hspace{0.1cm}A \hspace{0.1cm}\textsf{is any constant.} \\ \textsf{The solution of this ODE passes through}\hspace{0.1cm} (0, A)\\\textsf{and will never pass through}\hspace{0.1cm}(0, 0) \hspace{0.1cm}\textsf{except when} \hspace{0.1cm} A = 0. \\ \textsf{Since}\hspace{0.1cm} A\hspace{0.1cm} \textsf{can assume any value,}\\\hspace{0.1cm}(0, A) \hspace{0.1cm}\textsf{is not unique.}\\ \textsf{Therefore, we can conclude that}\\ \textsf{the solution of the differential equation}\\ \frac{\mathrm{d}y}{\mathrm{d}x} =y \hspace{0.1cm}\textsf{with}\hspace{0.1cm} y(0)=0\hspace{0.1cm} \textsf{exists, but is not unique.}$