Suppose P(x) is continuous on some interval I and a is a number in I. What can be said about the solution of the initial-value problem \f$y'+P(x)y=0, \ y(a)=0\f$ ?

As always, first,  I would like to write down the given data.

\f$\displaystyle y'+P(x)y=0, \ y(a)=0\f$

Sometimes, we have to rearrange the differential equation into the standard from but in this case, he differential equation is already in the standard form and so, we can start solving the problem.

I see that integration factor is  \f$e^{\int Pdx}\f$ and multiplying both sides of the equation by  \f$e^{\int P(x)dx}\f$ gives the following.

\f$\displaystyle \frac{d}{dx}(e^{\int P(x)dx}y)=0\f$

Integrate both side of the equation

\f$\displaystyle e^{\int P(x)dx}y=C\f$

Divide both side by  \f$e^{\int P(x)dx}\f$ to solve for y

\f$\displaystyle y=Ce^{-\int Pdx}\f$

Let’s remember this equation because I will come back to it and substitute the value of y into it.

In the question, it was given that  \f$y(a)=0\f$ , which means that y is equal to 0 when x is equal to a. By knowing this, we can figure out the following.

\f$\displaystyle y=Ce^{-\int Pdx}\f$

\f$\displaystyle C=0\f$

Therefore, we can said the following about the solution of the initial value problem  \f$y'+P(x)y=0, \ y(a)=0\f$ .

\f$\displaystyle y=0\f$  is a solution of the given initial-value problem and it is unique.

I hope my explanation makes sense and the solution is correct. If not, please do not hesitate to correct me, as I can benefit from my mistake and will not make the same mistake when I take a quiz this week.