 # if f(x) + x2[f(x)]5 = 34 and f(1) = 2, find f ‘(1). 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,  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.

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 $f'(x)$.

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

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

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

$f'(1) + 2(32) + 5(16) f'(1) = 0$

$f'(1) + 64 + 80f'(1) = 0$

$81f'(1) = -64$

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

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.