A differential equation refers to the mathematical equation that relates some function with its corresponding derivatives. The functions represent some physical quantities in real-life applications, while the derivatives represent the rate of change of the function concerning its independent variables.

In two variables, the most general differential equation is of the form -

f(p, q, q’, q”......) = c,

where

f(p, q, q’, q”...) is a function of p, q, q’, q”... and so on

p is the independent variable.

q is the dependent variable

q’, q” are the first-order and the second-order derivative of q respectively

c is some constant

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 that 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

• There must be no involvement of the 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, then that means it either has no degree or has an undefined degree.

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.

A first order differential equation is linear, when there is only dy/dx and not d2y/dx2, d3y/dx3 and so on, and can be made to look like:

dy/dx + P(x)y = Q(x), where 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 the equation 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)

Steps

1. Substitute y = uv

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:

Step 1

Substitute y = uv, and dy/dx = u dv/dx + v du/dx

So, the equation becomes – udv/dx + vdu/dx – uv/x = 1

Step 2

Factor the parts involving v

udv/dx + v( du/dx – u/x ) = 1

Step 3

Equate the v term with 0

du/dx – u/x = 0

so, du/dx = u/x

Step 4

To find u, solve using separation of variables

Separate variables: du/u = dx/x

Put integral sign: ∫ du/u = ∫ dx/x

Integrate: ln(u) = ln(x) + C

Make C = ln(k): ln(u) = ln(x) + ln(k)

So, u = kx

Step 5

Substitute u back into the equation we got at step 2

kx dv/dx = 1

Step 6

Solve the equation to find v

Separate variables: k dv = dx/x

Put integral sign: ∫ k dv = ∫ dx/x

Integrate: kv = ln(x) + C

Make C = ln(c): kv = ln(x) + ln(c)

So, kv = ln(cx)

And, v = 1/k ln(cx)

Step 7

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)

Let us consider a differential equation of the type y′′+py′+qy=0, where p,q are some constant coefficients.

For each of the equation, we can write the characteristic or auxiliary equation, which is of the form:

k^{2}+pk+q=0.

The general solution of the homogeneous differential equation depends on the roots of the characteristic quadratic equation. There are 3 different cases, which are as follows:

1. Discriminant or D of the characteristic quadratic equation is greater than 0, i.e., D>0. Then, the roots of the characteristic equations k1 and k2 are real and distinct. In this case, the general solution is given by the following function:

y(x)= C1ek1x + C2ek2x

where C1 and C2 are arbitrary real numbers.

2. Discriminant or D of the characteristic quadratic equation is equal to 0, i.e., D=0. Then, the roots are real and equal. It is said in this case that there exists one repeated root k1 of order 2. The general solution of the differential equation has the following form:

y(x)=(C1x+C2)ek1x

3. Discriminant or D of the characteristic quadratic equation is less than 0, i.e., D<0. Such an equation has complex roots k1=α+βi, k2=α−βi. The general solution is written as:

y(x)=e^αx *C1cos(βx)+C2sin(βx).