Second-Order Linear Differential Equations with Variable Coefficients

The second-order linear differential equations with variable coefficients are differential equations whose coefficients are a function of a certain variable. A second-order linear differential equation has a general form

where P, Q and R are functions of the independent variable x. If P and Q are some constant quantities, then the above equation is known as a second-order linear differential equation with constant coefficients.

If R = 0 then the equation is called a homogeneous linear differential equation of second order, otherwise it is non-homogenous.

Definition

Given real-valued functions a_o, a₁ and b such that a_o(t), a₁(t) and b(t) ∈ R ∀ t ∈ R, the differential equation of the form

y” + a₁(t) y’ + a_o(t) y = b(t)

is known as a second-order linear differential equation with variable coefficients. The variable coefficients are a_o(t) and a₁(t). If b(t) = 0 then the above equation is called a homogeneous second-order differential equation.

Here,

For example, y” + 2y’ + 6 = 0 is a second-order linear differential equation with constant coefficient.

y” + 2t y’ + log_e t y = e^3t is a second-order differential equation with variable coefficients.

Solution of Second-Order Differential Equations with Variable Coefficients

The solution of the second-order linear differential equation with variable coefficients can be determined using the Laplace transform. In particular, when the equations have terms of the form t^my⁽ⁿ⁾(t), its Laplace transform is (– 1)^m d^m/ds[L{y⁽ⁿ⁾(t)}].

Let us understand with an example, we have a second-order linear differential equation with variable coefficients

ty’’ + (1 – 2t)y’ – 2y = 0 where y(0) = 1 and y'(0) = 2

Taking Laplace transforms on both the side, we have

L{ty”} + L{y’} – 2L{ty’} – 2L{y} = 0

Integrating both sides, we get

ln|z| + ln|s – 2| = ln C₁

Or, L{y} = C₁/(s – 2)

Now, taking the inverse Laplace transform on both sides, we get

y = C₁ L^-1{1/(s – 2)} = C₁ e^2t

⇒ y = C₁ e^2t

But y(0) = 1, therefore C₁ = 1

Hence, the solution of the given differential equation is y = e^2t.

General and Fundamental Solutions

Before defining the Fundamental and general solution of a second-order linear differential equation with variable coefficients, we must know about the Wronskian of functions.

The Wronskian of function y₁, y₂: R → R is the function defined by

W_y1y2(t) = y₁(t) y₂‘(t) – y₂(t) y₁‘(t) ∀ t ∈ R

Alternatively,

Fundamental Solution

If y₁ and y₂ are two solutions of the differential equation y” + a₁(t) y’ + a_o(t) y = 0, then y₁ and y₂ are called fundamental solution if and only if y₁ and y₂ are linearly independent, that is, W_y1y2 ≠ 0.

General Solution

If y₁ and y₂ are two fundamental solution of the differential equation y” + a₁(t) y’ + a_o(t) y = 0, and c₁ and c₂ be any two arbitrary constants, then

y(t) = c₁ y₁(t) + c₂ y₂(t)

is said to be the general solution of the given differential equation.

Related Articles

• Differential Equations
• Applications of Differential Equations
• Ordinary Differential Equations
• Partial Differential Equations
• Frobenius Method

Solved Examples

Example 1:

Solve y” – ty’ + y = 1 if y(0) = 1 and y'(0) = 2.

Solution:

Taking Laplace transform on both sides of the given differential equation

L{y”} – L{ty’} + L{y} = L{1}

⇒ s²L{y} – sy(0) – y'(0) + d/ds[L{y’}] + L{y} = 1/s

⇒ s²L{y} – s – 2 + d/ds[sL{y} – y(0)] + L{y} = 1/s

⇒ s²z – s – 2 + d/ds[sz – 1] + z = 1/s where z = L{y}

⇒ s(dz/ds) + (s² + 2) z = s + 2 + 1/s

⇒ dz/ds + (s + 2/s) z = 1 + 2/s + 1/s² …….(1)

which is a linear differential equatin in z

Therefore,

The solution of (1) is

Putting s²/2 = u so that s ds = du

Or,

Or,

Now taking Inverse Laplace transform on both sides, we get

y = (2 + C₁)t + 1, since L^–1 {sⁿ} = 0 for n = 0, 1, 2, 3, …

But y'(0) = 2; 2 = C₁ + 2 ⇒ C₁ = 0

Hence the solution of the given differential equation is y = 2t + 1.

Example 2:

Show that y₁ = √t and y₂ = 1/t are fundamental solution of the differential equation

2t² y” + 3t y’ – y = 0.

Solution:

We show that W_y1y2 ≠ 0

y₁(t) = √t and y₁‘(t) = ½(t ^{– ½} )

y₂ = 1/t and y₂‘(t) = – t ^{– 2}

Now, W_y1y2 = t^1/2 (– t ^{– 2}) – 1/t [ ½(t ^{– ½} )]

= – 3/2 (t ^{– 3/2}) ≠ 0

Therefore, y₁ and y₂ are fundamental solution of the given differential equation.