# Differential Equations

### Summary

• Differential Equation – any equation which involves $\frac { dy }{ dx }$ or any higher derivative.
• Solving differential equations means finding a relation between y and x alone through integration.
• We use the method of separating variables in order to solve linear differential equations.
• We must be able to form a differential equation from the given information.

A differential equation is just an equation involving a function and its derivatives. In other words any equation which involves $\frac { dy }{ dx }$ or any higher derivative is known as a “Differential Equation”.

Solving a differential equation means finding the functions itself through integration.

We can either classify differential equations as first order, second order or higher. I.e: $2\frac { dy }{ dx } \quad =\quad xy$  is a first order differential equation (also called linear differential equation) $\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \quad +\quad 2\frac { dy }{ dx } \quad +\quad 3y\quad =\quad 8$  is a second order differential equation.

Solving differential equations involves integration. We basically find a relation between y and x alone.

We use the method of separating variables in order to solve linear differential equations.

The method works in the following way:

• E.g: we have a differential equation of first order: $\frac { dy }{ dx } \quad =\quad f\left( x \right)g\left( x \right)$

• We now separate both variables x and y: $\frac { dy }{ g\left( y \right) } \quad =\quad f\left( x \right) dx$

• We then integrate both sides of the equation: $\int { \frac { dy }{ g\left( y \right) } } \quad =\quad \int { f\left( x \right) dx }$

• After both sides are integrated we get a function in terms of x and y only, and we must add a constant on one side of the equation when no limits or boundaries are given. $G(y)\quad =\quad F(x)\quad +\quad C$

For you to understand this method better, let’s go through a few examples.

#### Example #1

Q. Solve the differential equation $y\quad \frac { dy }{ dx } \quad =\quad x\quad { e }^{ x }$

Solution: $y\quad \frac { dy }{ dx } \quad =\quad x\quad { e }^{ x }$

Separate the variables: $y\quad dy\quad =\quad x\quad { e }^{ x }dx$

Integrate both sides: $\int { y\quad dy } \quad =\quad \int { x\quad { e }^{ x }dx }$

Integration by parts of RHS: $\frac { { y }^{ 2 } }{ 2 } \quad =\quad x{ e }^{ x }\quad -\quad \int { x{ e }^{ x }(1)\quad dx }$ $\frac { { y }^{ 2 } }{ 2 } \quad =\quad x{ e }^{ x }\quad -\quad { e }^{ x }\quad +\quad C$

(add a constant since no limits are given)

Equation becomes:

Ans: ${ y }^{ 2 }\quad =\quad 2\left[ x{ e }^{ x }\quad -\quad { e }^{ x }\quad +\quad C \right]$

#### Formation of Differential equations

Now that you understand how to solve a given linear differential equation, you must also know how to form one. In many scenarios we will be given some information, and the examiner will expect us to extract data from the given information and form a differential equation before solving it.

Mostly scenarios, involve investigations where it appears that the rate at which one variable is changing with respect to another variable, can be expressed by a simple differential equation.

A differential equation depending on the data is formed and then its possible solutions are worked out through integration.

A very common scenario is that of a “ Chemical Reaction” as shown in the example below.

#### Example #2

Q. In a certain industrial process, a substance is being produced in a container. The mass of the substance in the container t minutes after the start of the process is x grams. At any time, the rate of formation of a substance is proportional to its mass.

Also throughout the process , the substance is removed from the container at a constant rate of 25 grams/minute when t = 0, x = 1000 and $\frac { dx }{ dt } \quad =\quad 75$.

Form a differential equation and solve it to obtain an expression for x in terms of t.

Solution:

Let x = Amount of the substance at any time t

Rate of formation is said to be change in the amount of substance with respect to t which is $\frac { dx }{ dt }$.

We can say that:

Rate of change of substance = Rate of formation – Rate of removal of substance $\frac { dx }{ dt } \quad =\quad kx\quad -\quad 25\quad \quad \quad \quad \quad \quad \quad \quad =>\quad eq\quad 1$

Substitute x = 1000, and $\frac { dx }{ dt } \quad =\quad 75$ to find k: $75\quad =\quad k(1000)\quad -\quad 25$ $k\quad =\quad 0.1$

Equation 1 becomes: $\frac { dx }{ dt } \quad =\quad (0.1)x\quad -\quad 25$

(differential equation formed)

Now solve the equation.

Write equation in this form: $\frac { dx }{ dt } \quad =\quad \frac { 1 }{ 10 } (x\quad -\quad 250)$

Seperate variables: $\frac { dx }{ x\quad -\quad 250 } \quad =\quad \frac { 1 }{ 10 } dt$

Integrate : $\int { \frac { dx }{ x\quad -\quad 250 } } \quad =\quad \int { \frac { 1 }{ 10 } dt }$ $\ln { \left| x\quad -\quad 250 \right| } =\quad 0.1t\quad +\quad \ln { C } \quad \quad \quad \quad \quad \quad \quad \quad =>\quad \quad eq\quad 2$

Find the value of constant since we know the value x = 1000 at t = 0: $\ln { \left| 1000\quad -\quad 250 \right| } =\quad 0.1(0)\quad +\quad \ln { C }$ $C\quad =\quad 750$

Substitute the value of C in equation 2 to obtain an expression for x in terms of t $\ln { \left| x\quad -\quad 250 \right| } =\quad 0.1t\quad +\quad \ln { (750) }$  $\ln { \left| x\quad -\quad 250 \right| } =\quad \ln { { e }^{ 0.1t } } \quad +\quad \ln { (750) }$ $\ln { \left| x\quad -\quad 250 \right| } =\quad \ln { (750\quad { e }^{ 0.1t }) }$ $\quad x\quad -\quad 250\quad =\quad 750\quad { e }^{ 0.1t }$

Ans: $\quad x\quad =\quad 750\quad { e }^{ 0.1t }\quad +\quad 250$