Q. 6 - Calculus

Question

Initial-Value Problems: An initial-value problem is a differential equation together with an initial condition that specifies one value of a function.

• Find the solution $f$ to the initial-value problem $f^{\text{'}}\left(x\right)=$ $\sin x, f\left(\pi \right)=0$.

Step-by-Step Solution

Answer

Ans: $f\left(x\right)=-\cos x-1$

Step 1. Given information:

The initial-value problem $f^{\text{'}}\left(x\right)=\sin x,f\left(\pi \right)=0.$

Step 2. Solving the differential equation :

In order to solve the given differential equation $f^{\text{'}}\left(x\right)=\sin x$, since the differential equation is already separable, so integrate both sides with respect to $x$ to get

$$\begin{gathered} \int f^{\text{'}}\left(x\right)=\int \sin xdx \\ f\left(x\right)=-\cos x+C \end{gathered}$$

Here $\mathrm{C}$ is the integration constant.

Now use the initial condition $f\left(\pi \right)=0$ and simplify further for C to get

$$\begin{gathered} f\left(\pi \right)=-\cos \left(\pi \right)+C \\ 0=-(-1)+C \\ 0=1+C \\ \Rightarrow C=-1 \end{gathered}$$

Thus, the solution of the given differential equation is $f\left(x\right)=-\cos x-1$.