From y=c1+c2 x^2 we find y^'=2 c2 x . Then y(0)=c1=0, y^'(0)=2 c2 ·0=0 and hence y^'(0)=1 is not

step 1 y of 0 is equal to 0 gives c1 plus c2 of 0 square that is equal to 0 now this equation implies t

From y=c1+c2 x^2 we find y^'=2 c2 x . Then y(0)=c1=0,...

From $y=c_{1}+c_{2} x^{2}$ we find $y^{\prime}=2 c_{2} x .$ Then $y(0)=c_{1}=0, y^{\prime}(0)=2 c_{2} \cdot 0=0$ and hence $y^{\prime}(0)=1$ is not possible. since $a_{2}(x)=x$ is 0 at $x=0,$ Theorem 3.1 is not violated.

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Key Concepts

Existence and Uniqueness Theorem

The existence and uniqueness theorem provides conditions that guarantee a differential equation has one and only one solution that satisfies given initial conditions. A key requirement is often that certain coefficient functions do not vanish in the region of interest; if they do, as in the example where a coefficient becomes zero, the theorem’s conditions may not be applicable or may need modification.

Coefficient Functions in Differential Equations

In a differential equation, the coefficients, particularly those multiplying the derivative terms, play an essential role in the behavior of the solution. If a coefficient (like the one multiplying a derivative) vanishes at a point, it can indicate a singular or degenerate situation, requiring careful analysis of solution existence and uniqueness.

Initial Value Problems

An initial value problem specifies the value of the unknown function and possibly some of its derivatives at a particular point. This setup is critical in determining whether a differential equation has a unique solution that satisfies these specified initial conditions.

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