Chapter 3, Problem 1CR

To determine

To fill: The blank in the statement, “The only solution of the initial-value problem $y^{″}+x^{2}y=0,y\left(0\right)=0,y^{\prime }\left(0\right)=0$ is ______.”

Answer to Problem 1CR

The only solution of the initial value problem $y^{″}+x^{2}y=0,y\left(0\right)=0,y^{\prime }\left(0\right)=0$ is $y\left(x\right)=0$.

Explanation of Solution

Theorem used:

The existence of a Unique Solution:

For a n^th order initial value problem $a_n{y}^{\left(n\right)}+\mathrm{...}+a_1{y}^{\prime }+a_{0}y=g\left(x\right)$, where $a_n,a_{n-1},\mathrm{...},a_0\text{ and }g\left(x\right)$ are continuous on an interval I with $a_n\left(x\right)\ne 0$ on I.

If $x=x_0$ is a point in interval I, then a solution $y\left(x\right)$ of initial value problem exists on I and the solution is unique.

Calculation:

Consider the initial value problem $y^{″}+x^{2}y=0,y\left(0\right)=0,y^{\prime }\left(0\right)=0$. The differential equation $y^{″}+x^{2}y=0$ is a second order differential equation.

Compare the equation to the standard form of differential equation $a_n{y}^{\left(n\right)}+\mathrm{...}+a_1{y}^{\prime }+a_{0}y=g\left(x\right)$. Here, $a_2\left(x\right)=1,a_1\left(x\right)=0,a_0\left(x\right)=x^2\text{ and }g\left(x\right)=0$.

Note that, $a_2\left(x\right)=1,a_1\left(x\right)=0,a_0\left(x\right)=x^2\text{ and }g\left(x\right)=0$ are continuous on $I=\left(-\infty ,\infty \right)\text{ and }{x}_0=\left(0\right)\in \left(-\infty ,\infty \right)$.

The initial value $x_0=0$ is in the interval I and it satisfies the initial value problem. Thus the unique solution of the initial value problem is $y\left(x\right)=0$.

By the existence of unique solution, the general solution of the initial value problem $y^{″}+x^{2}y=0$ with initial values $y\left(0\right)=0,y^{\prime }\left(0\right)=0$ has a unique solution $y\left(x\right)=0$.

Therefore, the only solution of the initial value problem $y^{″}+x^{2}y=0,y\left(0\right)=0,y^{\prime }\left(0\right)=0$ is $y\left(x\right)=0$.