 # Find the particular solution that satisfies the differential equation and the initial condition. f”(x) = sin (x) , f'(0) = 1,  f(0) = 6

This problem aims to familiarize us with the concepts of initial value problems. The concepts required to solve this problem are related to the basics of differential equations, which include the order of a differential equation, general and particular solutions, and initial value problems.

So a differential equation is an equation about an unspecified function y = f(x) and a series of its derivates. Now the particular solution to a differential is a function y = f(x) that fulfills the differential when f and its derivatives are plugged into the equation, whereas the order of a differential equation is the highest ranking of any derivative that occurs in the equation.

We know that any solution of a differential equation is of the form $y=mx + C$. This is an illustration of a general solution. If we find the value of $C$, then it is known as a particular solution to the differential equation. This particular solution can be a unique identifier if some additional information is given.

So, let’s first integrate the double derivative to simplify it into a first derivative:

$f^{”}(x)=\sin (x)$

$\int f^{”} dx=\int\sin x dx$

The first derivative of $\sin x$ is negative of $\cos x$:

$f'(x)=-\cos x+C_1$

Here, we get a constant $C_1$, which can be found using the initial condition given in the question $f'(0) = 1$.

Plugging in the initial condition:

$-\cos x+C_1=1$

$-1 + C_1=1$

$C_1=1+1$

$C_1=2$

So, the particular solution in the form of the first derivative comes out to be:

$f'(x)=\cos x+2$

Now, let’s integrate the first derivative to get the actual function:

$\int f'(x) dx=\int (-\cos x+2)dx$

$f(x)=\int -\cos x dx+\int 2 dx$

The first derivative of $cosx$ is equal to $sinx$:

$f(x)=-\sin x +2x+C_2$

Here, we get a constant $C_2$ which can be found using the initial condition given in the question $f(0)=6$.

Plugging in the initial condition:

$-\sin (0) + 2(0) +C_2 = 6$

$0 + C_2 = 6$

$C_2 = 6$

Finally, the particular solution of the given differential equation comes out to be:

$f(x) = -\sin x + 2x + 6$

## Numerical Result

The particular solution of the given differential equation comes out to be $f(x) = -\sin x + 2x + 6$.

## Example

Find the solution to the following initial-value problem:

$y'(x) = 3e^x + x^2 – 4,\space y(0) = 5$

The first step is to find a general solution. To do this, we find the integral of both sides.

$\int y'(x) dx =\int (3e^x + x^2 – 4) dx$

$y(x) + C_1 = 3e^x +\dfrac{1}{3}x^3 – 4x + C_2$

Note that we get two integration constants: $C_1$ and $C_2$.

Solving for $y$ gives:

$y(x) = 3e^x +\dfrac{1}{3}x^3 – 4x + C_2 – C_1$

Defining $C = C_2 – C_1$, since both are constant and will yield a constant:

$y(x) = 3e^x +\dfrac{1}{3}x^3 – 4x + C$

Substituting the initial condition:

$5=3e^0 +\dfrac{1}{3}0^3 – 40 + C$

$5=3+C$

$C=2$

$y(x) = 3e^x +\dfrac{1}{3}x^3 – 4x +2$