** 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.

## Expert Answer

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:**

\[5=3e^0 +\dfrac{1}{3}0^3 – 40 + C\]

\[5=3+C\]

\[C=2\]

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