Differential equations can be categorized into six different types, namely, ordinary differential equations, partial differential equations, linear differential equations, non-linear differential equations, homogeneous differential equations, and non-homogeneous differential equations.
For finding the solution of a differential equation, we can use two methods to solve the differential function, namely separation of variables and integrating factor.
Separation of variables is a technique implemented when the differential equation can be expressed or written in the form of dy/dx = p(y)q(x),
where, p is the function of y only, and q is the function of x only.
By taking an initial condition, we can rewrite the equation as 1/p(y)dy = q(x)dx and then apply integration on both sides.
The integrating factor technique is implemented when the differential equation is of the form dy/dx + p(x)y = q(x).
In this form, both p and q are functions of y only.
A first-order differential equation is linear, when there is only dy/dx and not d2y/dx2 or d3y/dx3 or any other derivative. It looks like:
dy/dx + P(x)y = Q(x)
In this equation, P(x) and Q(x) are functions of x.
There’s a special method for solving this equation.
We use two new functions of x, let them be u and v, and say y=uv.
Then, we will solve to find u and v.
Also, we will find the derivative of y=uv, using the product rule.
The derivative, dy/dx = udv/dx + vdu/dx (differentiating with respect to x)
1. Start by substituting y = uv, and dy/dx = udv/dx + vdu/dx into the equation dy/dx + P(x)y = Q(x)
2. Factor the parts that involve v
3. Equate the v term with zero and solve using separation of variables to find u
4. Substitute u into the equation we got at step 2
5. Solve the equation to find v
6. Substitute u and v into the equation y=uv to find the final answer
Let’s understand this using an example
dy/dx – y/x = 1
The given differential equation is linear, so let’s follow the steps mentioned above:
Start by substituting y = uv, and dy/dx = u dv/dx + v du/dx
So, the equation becomes – udv/dx + vdu/dx – uv/x = 1
Factor the parts involving v
udv/dx + v( du/dx – u/x ) = 1
Equate the v term with 0
du/dx – u/x = 0
so, du/dx = u/x
To find u, solve using separation of variables
Separate variables: du/u = dx/x
Now, put the integral sign: ∫ du/u = ∫ dx/x
Integrate: ln(u) = ln(x) + C
Next, make C as ln(k): ln(u) = ln(x) + ln(k)
So, u = kx
Substitute the value of u back into the equation that we got at step 2
kx dv/dx = 1
Solve the equation to find v
Separate variables: k dv = dx/x
Now, put the integral sign: ∫ k dv = ∫ dx/x
Integrate: kv = ln(x) + C
Next, make C as ln(c): kv = ln(x) + ln(c)
So, kv = ln(cx)
And, v = 1/k ln(cx)
Now, substitute the values in y = uv, to find the final answers for the original equation
y = kx 1/k ln(cx) and simplify
So, the answer is y = x ln(cx)
The order of a differential equation refers to the highest order derivative involved in that particular differential equation. The degree of a differential equation refers to the exponent or the power of the highest order derivative involved in that particular differential equation, provided the differential equation satisfies the conditions specified below:
The derivatives in the equation must be free from both negative and positive fractional powers if any.
No involvement of derivatives in any fraction.
There must not be any involvement of the highest order derivative as a transcendental, exponential, or trigonometric function. The coefficient of any term in the differential equation containing the highest order derivative should only be a function of p, q, or some lower-order derivative.
If one or more of the conditions mentioned above are not satisfied by the differential equation, then it first needs to be reduced to the form in which it satisfies all of the conditions. In case the equation isn't reducible, it has no degree or has an undefined degree.
Let’s consider a few examples:
X d2y/dx2 +Y dy/dx + 4y2
The given differential equation is already in the reduced form. The highest order derivative in this equation is of order 2, and its power or exponent is 1. Therefore, the order of the differential equation is 2 and its degree is 1.
3y2(dy/dx)3 - d2y/dx2=sin(x/2)
The highest order derivative involved in this particular differential equation, which is already in the reduced form, is of order 2 and its corresponding power is 1. Therefore, the order of the differential equation is 2 and its degree is 1.
An ordinary differential equation (ODE) consists of a function and its derivatives. It has only a single independent variable along with one or more of its derivative w.r.t the variable. The order of an ODE is the order of the highest derivative occurring in the equation.
Differential equations are extensively used in describing several exponential growths and decays.
In the field of medicines, differential equations help in modeling cancer growth or the spread of other diseases in the human body.
They are widely used in describing the change in investment return over time.
Economists use differential equations to find optimum investment strategies.
These equations also help in describing the movement of electricity and the motion of waves.
For understanding the real-life application of a linear differential equation, let us consider the example of exponential growth (population). Let P(t) denote a quantity that increases with time t, and the rate of increase of population is proportional to the same quantity P, expressed as follows:
dp/dt = kP, where dp/dt is the first-order derivative of P w.r.t to time t, k is the proportionality constant, and k>0.
The solution to the above equation is P(t) = Po ekt (where Po is constant and not equal to 0).
Since k is positive and assuming that Po is also positive, we can say that P(t) is an increasing exponential. Hence, dp/dt = kP is also known as the exponential growth model.