Monday, December 31

Solving Simple Differential Equation


Solving simple differential equations involve the process of differentiating the algebraic function with respect to the input function. The algebraic function which is differentiable is known as differential equations. The differential equation comes under calculus category whereas to find the rate of change of the given function with respect to the input function. The following are simple example differential equations for solving.


I like to share this Balancing Equations with you all through my article.

Simple Differential Equations Examples for Solving:

The following are the example problems with simple differential equations for solving.

Example 1:

Solve the simple differential equation.

f(k) = k2 – 4k + 8

Solution:

The given equation is

f(k) = k 2 – 4k + 8

The first derivative f ' for the algebraic function is

f '(k) = 2 k  – 4

Example 2:

Solve the simple differential equation.

f(k) = k 3 – 5 k 2  + 11k

Solution:

The given function is

f(k) = k 3 – 5 k 2  + 11k

The first derivative f ' for the algebraic function is

f '(k) = 3k 2 – 5(2 k  ) + 11

f '(k) = 3k 2 – 10 k + 11

Example 3:

Solve the simple differential equation.

f(k) = k4 – 3k 3 – 4k 2  + k

Solution:

The given function is

f(k) = k4 – 3k 3 – 4 k 2  + k

The first derivative f ' for the algebraic function is

f '(k) = 4 k 3 – 3(3k 2 ) – 4( 2 k  ) + 1

f '(k) = 4 k 3 – 9k 2  – 8 k  + 1

My forthcoming post is on Example of Hypothesis Testing and Dividing Fraction will give you more understanding about Algebra.

Example 4:

Solve the simple differential equation.

f(k) = k 5 – 6 k 3  + 11

Solution:

The given function is

f(k) = k 5 – 6 k 3  + 10

The first derivative f ' for the algebraic function is

f '(k) = 5k 4 – 6(3 k 2 )

f '(k) = 5k 4 – 18 k 2
Simple Differential Practice Equations for Solving:

1) Solve the simple differential equation.

f(k) = k 3 – 6 k 2  + 11k

Answer: f '(k) = 3k 2 – 12 k

2) Solve the simple differential equation.

f(k) = k 2 – 6 k   + 11

Answer: f '(k) = 2k – 6

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