This question belongs to the **calculus** domain and **aims** to explain the **differential** equations and **initial** value problems.

In Calculus, a **differential equation** is an equation that includes one or more **functions** with their **derivatives.** The rate of change of a **function** at a point is defined by the function’s **derivatives.** It is **primarily** used in fields like physics, biology, engineering,Â etc. The preliminary **objective** of the differential **equation** is to **analyze** the solutions that benefit the **equations** and the **properties** of the solutions.

A **differential** equation holds **derivatives** that are either **ordinary** derivatives or **partial** derivatives. The **derivative** conveys the rate of **change,** and the **differential** equation defines a **connection** between the quantity that is **continuously** altering with respect to the **transition** in another quantity.

An **initial value** problem is a **standard** differential **equation** jointly with an **initial** condition that **specifies** the value of the **unspecified** function at a **provided** point in the **domain.** Modeling a system in **physics** or other sciences often **amounts** to solving an **initial** value problem.Â

## Expert Answer

Given **Function:**

\[ f(x) + x^2[f(x)]^5 = 32 \]

Given the **value** of function:

\[ f(1) = 2 \]

And we have to **find** $f'(1)$.

In the first step, Apply the **differentiation** with respect to $y$ on the given **equation:**

\[ \dfrac{d}{dy} (f(x) + x^2[f(x)]^5 = 32) \]

\[ \dfrac{d}{dy} f(x) + \dfrac{d}{dy} (x^2[f(x)]^5) = \dfrac{d}{dy} (32) \]

\[ f'(x) + [f(x)]^5 \dfrac{d}{dx}x^2 + x^2 \dfrac{d}{dx}[f(x)]^5 = 0 \]

\[ f'(x) + 2x[f(x)]^5 + x^2 \times 5 \times [f(x)]^{5-1} \dfrac{d}{dx}[f(x)] = 0 \]

\[ f'(x) + 2x[f(x)]^5 + x^2 \times 5 \times [f(x)]^{4} f'(x) = 0 \]

Now putting the **given** information $f(1)=2$ and **solving** $f'(x)$.

\[ f'(1) + 2(1)[f(1)]^5 + (1)^2 \times 5 \times [f(1)]^{4} f'(1) = 0 \]

\[ f'(1) + 2[2]^5 +Â 5 \times [2]^{4} f'(1) = 0 \]

\[ f'(1) + 2[2]^5 +Â 5 \times [2]^{4} f'(1) = 0 \]

\[ f'(1) + 2(32) +Â 5(16) f'(1) = 0 \]

\[ f'(1) + 64 +Â 80f'(1) = 0 \]

\[ 81f'(1) = -64 \]

\[ f'(1) = \dfrac{-64}{81} \]

## Numerical Answer

Given $f'(1) =2$ $f'(1)$ **comes** out to be $\dfrac{-64}{81}$

## Example

Show that the **function** $y=2e^{-2t} +e^t$ proves to the **initial-value** problem:

\[ y’ +2y = 3e^t, \space y(0)=3 \]

The initial-value problem is **satisfied** when both the **differential** equation and the **initial** condition **satisfy.** Starting the solution by **calculating** $y’$, to prove $y$ satisfies the **differential** equation.

\[ y=2e^{-2t} +e^t \]

\[ \dfrac{d}{dt}y=\dfrac{d}{dt}(2e^{-2t} +e^t) \]

\[ y’=\dfrac{d}{dt}2e^{-2t} +\dfrac{d}{dt}e^t \]

\[ y’= -2(2e^{-2t})\dfrac{d}{dt}(t) +e^t \]

\[ y’= -4e^{-2t} +e^t \]

Next, we **replace** both $y$ and $y’$ into the **left-hand** side of the differential **equation** and solve:

\[ y’ +2y = (-4e^{-2t}) +e^t) + 2(2e^{-2t} +e^t) \]

\[ = -4e^{-2t}) +e^t + 4e^{-2t} + 2e^t \]

\[ 3e^t \]

That is equals to the **right** hand side of the differential equation, $y= 2e^{-2t} +e^t$ proves the **differential** equation. Next we find $y(0)$:

\[y(0)=2e^{-2(0)}+e^0 \]

\[y(0)=3\]

The given function **proves** the initial value problem.