Question

Suppose $P\left(x\right)$is continuous on some interval I and ais the number in I. What can be said about the solution of the initial-value problem

$y\text{'}+P\left(x\right)y=0,y\left(a\right)=0?$

Short answer

The possible solution of initial value problem is y = 0.

Step-by-step solution

Definition of initial-value problem

The interval I, containing the problem solving an nth order differential equations subjected to n side conditions specified at x₀

Solve:

$\begin{array}{l}\frac{d^{n}y}{d{x}^n}=F(x,y,y\text{'},.....,y^{(n-1)})\\ y\left(x_0\right)=y_0,y\text{'}\left(x_0\right)=y_1,...,y^{(n-1)}\left(x_0\right)=y_{n-1}\end{array}$

Where, $y_0,y_1,.....,y_{n-1}$are arbituary constants called an nth-order initial value problem (IVP)

The values of y(x) and its first (n-1) derivatives at $y_0,y\text{'}\left(x_0\right)=y_1,...,y^{(n-1)}\left(x_0\right)=y_{n-1}$are called initial conditions(IC)

Determination of initial value problem

The given initial value problem is $y\text{'}+P\left(x\right)y=0$where

The given initial-value problem is in standard form

The integrating factor will be

$e^{\int P\left(x\right)dx}$

Now by using the integrating factor we will multiply the given equation .

we get ;

$\begin{array}{l}{e}^{\int P\left(x\right)dx}y\text{'}+P\left(x\right)y{e}^{\int P\left(x\right)dx}=e^{\int P\left(x\right)dx}.0\\ \frac{d}{dx}\left[e^{\int P\left(x\right)dx}y\right]=0\\ \int \frac{d}{dx}\left[e^{\int P\left(x\right)dx}y\right]dx=\int 0 dx\\ e^{\int P\left(x\right)dx}y=C\\ y=C{e}^{-\int P\left(x\right)dx}\end{array}$

The given initial condition is $y\left(a\right)=0andP\left(x\right)$is continuous

is continuous on interval of a lies, where c=0

The c= 0, so there is only possible solution will be y =0