Question $4$

$0\mathrm{.}4$ pts

Determine whether the Existence and Uniqueness Theorem implies that the given initial value problem has a unique solution.

$y^3\frac{dy}{dx}=x^7,y(6)=0$

$14$

The theorem does not imply the existence of a unique solution because $\frac{x^7}{y^3}$ is not continuous in any rectangle containing the point $(6,0).$

The theorem does not imply the existence of a unique solution because $\frac{x^7}{y^3}$ is continuous but $\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{y}}(\frac{x^7}{y^3})$ is not continuous in any rectangle containing the point $(0,6).$

The theorem does not imply the existence of a unique solution because $\frac{x^7}{y^3}$ is continuous but $\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{y}}(\frac{x^7}{y^3})$ is not continuous in any rectangle containing the point $(6,0).$

The theorem implies the existence of a unique solution beqause $\frac{x^7}{y^3}$ and $\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{y}}(\frac{x^7}{y^3})$ are both continuous in any rectangle containing the point $(0,6).$