13. High order Linear Differential Equations

Dated: 15-11-2024

The equations of the form

\[ a_{n}(x)\frac{d^{n}y}{dx^{n}}+a_{n-1}(x)\frac{d^{n-1}y}{dx^{n-1}}+\cdot\cdot\cdot+a_{1}(x)\frac{dy}{dx}+a_{0}(x)y=g(x) \]

or

\[ a_{n}(x)y^{(n)}+a_{n-1}(x)y^{(n-1)}+\cdot\cdot\cdot+a_{1}(x)y^\prime+a_{0}(x)y=g(x) \]

where \(a_0(x), a_1(x), \ldots, a_n(x), g(x)\) are functions,¹ are called higher order linear differential equations with variable coefficients.
First, we will study the ones with constant coefficients which look like

\[ a_{n}\frac{d^{n}y}{dx^{n}}+a_{n-1}\frac{d^{n-1}y}{dx^{n-1}}+\cdot\cdot\cdot+a_{1}\frac{dy}{dx}+a_{0}y=g(x) \]

if \(g(x) = 0\) then this equation is called associated homogeneous differential equation.

Initial Value Problem

Solve

\[ a_{n}(x)\frac{d^{n}y}{dx^{n}}+a_{n-1}(x)\frac{d^{n-1}y}{dx^{n-1}}+\cdot\cdot\cdot+a_{1}(x)\frac{dy}{dx}+a_{0}(x)y=g(x) \]

Subject to the following arbitrary constants called the initial conditions

\[y(x_0) = y_0, y^\prime(x_0) = y^\prime_0, \ldots y^{n-1}(x) = y_0^{n - 1}\]

is called the initial value problem.

Solution

A function¹ satisfying the differential equation on \(I\) whose graph passes through \((x_0, y_0)\) such that the slope² of the curve at the point is the number \(y_0^\prime\) is called solution of the initial value problem.

Theorem

Let the variable coefficients be continuous³ over the interval \(I\) and let \(a_n(x) \ne 0, \forall x \in I\).
If \(x = x_0 \in I\) then a solution \(y(x)\) of the initial value problem exists on \(I\) and is unique.

Example

Consider the function¹

\[y = cx^2 + x + 3\]

With initial value problem

\[ x^2y^{\prime\prime}-2xy^{\prime}+2y=6 \]

\[y(0) = 3, y^\prime(0) = 1\]

Then

\[y^\prime = 2cx + 1\]

and

\[y^{\prime\prime} = 2c\]

Therefore,

\[ x^2y^{\prime\prime}-2xy^{\prime}+2y=x^2(2c)-2x(2cx+1)+2(cx^2+x+3) \]

\[ =2cx^2-4cx^2-2x+2cx^2+2x+6 \]

\[= 6\]

Also

\[y(0) = 3 \implies c(0) + 0 + 3 = 3\]

and

\[y^\prime(0) = 1 \implies 2c(0) + 1 = 1\]

So for any choice of \(c\), the \(y(x)\) satisfied the differential equation and initial conditions.
Therefore the solution to the equation is not unique.

Note that

• The equation is linear differential equation.⁴
• The coefficients being polynomials are continuous³ everywhere.
• The function¹ being constant is constant everywhere.
• The leading coefficient \(a_2(x) = x^2 = 0\) at \(x = 0 \in (-\infty, \infty)\).

Hence \(a_2(x)\) brought non uniqueness in the solution.

Boundary Value Problem (BVP)

For a 2nd order linear differential equation the problem of solving

\[ a_{2}(x)\frac{d^{2}y}{dx^{2}}+a_{1}(x)\frac{dy}{dx}+a_{0}(x)y=g(x) \]

subject to following boundary conditions

\[y(a) = y_0, \quad y(b) = y_1\]

is called a boundary value problem.

Solution

A solution of the boundary value problem is a function¹ satisfying the differential equation on some interval⁵ \(I\) , containing \(a\) and \(b\), whose graph passes through two points \((a, y_0)\) and \((b, y_1)\).

Example

Consider the function¹

\[y=3x^2-6x+3\]

We can prove that this function¹ is the solution to following boundary value problem

\[x^2y^{\prime\prime}-2xy^\prime+2y=6\]

\[y(1) = 0, y(2) = 3\]

Since

\[\frac {dy}{dx} = 6x - 6\]

and

\[\frac {d^2y}{dx^2} = 6\]

Therefore,

\[x^2\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+2y=6x^2-12x^2+12x+6x^2-12x+6=6\]

Also

\[y(1) = 3 - 6 + 3 = 0\]

\[y(2) = 12 - 12 + 3 = 3\]

Therefore, \(y(x)\) satisfies both, the differential equation and boundary conditions.

Possible Boundary Conditions

• \[y(a) = y_0, \quad y(b) = y_1\]
• \[y^\prime(a) = y^\prime_0, \quad y(b) = y_1\]
• \[y(a) = y_0, \quad y^\prime(b) = y^\prime_1\]
• \[y^\prime(a) = y^\prime_0, \quad y^\prime(b) = y^\prime_1\]

In General

The above possible boundary conditions are just a special case of the general boundary conditions

\[\alpha_1y(a) + \beta_1y^\prime(a) = \gamma_1\]

\[\alpha_2y(a) + \beta_2y^\prime(a) = \gamma_2\]

where

\[\alpha_1, \alpha_2, \beta_1, \beta_2 \in \{0, 1\}\]

A boundary value problem may have

Several Solutions
A unique solution
No solution at all

Example 1

Consider the function¹

\[y = c_1\cos (4x) + c_2 \sin (4x)\]

and the boundary value problem

\[y^{\prime\prime} + 16 y = 0, y(0) = 0, y\left(\frac \pi 2 \right) = 0\]

Then

\[y^{\prime} = -4c_1 \sin 4x + 4c_2 \cos 4x\]

\[y^{\prime\prime} = -16(c_1 \cos 4x + c_2 \sin 4x)\]

\[y^{\prime\prime} = -16y\]

\[y^{\prime\prime} + 16y = 0\]

Therefore, the function¹ satisfied the differential equation.
Apply the condition

\[y(0) = 0\]

We obtain

\[0 = c_1 \cos 0 + c_2 \sin 0\]

\[\implies c_1 = 0\]

So that

\[y = c_2 \sin (4x)\]

But when we apply the 2nd condition

\[y\left(\frac \pi 2\right) = 0\]

We have

\[0 = c_2 \sin (2 \pi)\]

\[\because \sin(2 \pi) = 0\]

The condition is satisfied for any choice of \(c_2\).
Solution is

\[y = c_2 \sin (4 \pi)\]

Hence there are an infinite number of solutions for the boundary value problem.

Example 2

Solve

\[y^{\prime\prime} + 16y = 0\]

\[y(0) = 0, y\left(\frac \pi 8\right) = 0\]

Solution

From the previous example, we know the following function¹ satisfied the problem

\[y = c_1\cos (4x) + c_2 \sin (4x)\]

Apply the conditions

\[y(0) = 0 \implies 0 = c_1 + 0\]

\[y\left(\frac \pi 8\right) = 0 \implies 0 = c_2 + 0\]

So

\[c_1 = c_2 = 0\]

Therefore, the only solution is

\[y = 0\]

Example 3

Same problem as previous one but with conditions

\[y(0) = 0, y \left(\frac \pi 2\right) = 1\]

For the \(y(0)\), it is same, \(c_1 = 0\).
So that

\[y = c_2 \sin(4x)\]

But

\[y\left(\frac \pi 2\right) = 1\]

\[\implies c_2 \sin(2 \pi) = 1\]

\[\because \sin(2 \pi) = 0\]

But

\[c_2 \cdot 0 = 1\]

which is a contradiction, hence there is no solution.

Linear Dependence

A set⁶ of functions¹

\[\{f_1(x), f_2(x), \ldots, f_n(x)\}\]

is said to be linearly dependent on an interval⁵ \(I\) if there exists constants \(c_1, c_2, \ldots, c_n\) not all zeros such that

\[c_1f_1(x)+c_2f_2(x)+\cdots+c_nf_n(x)=0, \quad \forall x \in I\]

Linear Independence

Same thing as linear dependence but only when

\[c_1 = c_2 = \cdots = c_n = 0\]

Case of Two Functions

If \(n = 2\) then the set⁶ of functions¹ becomes

\[\{f_1(x), f_2(x)\}\]

If we suppose that

\[c_1f_1(x)+c_2f_2(x)=0\]

and that the functions¹ are linearly dependent on an interval⁵ \(I\) then either \(c_1 \ne 0\) or \(c_2 \ne 0\).
Let us assume that \(c_1 \ne 0\) then

\[f_1(x) = - \frac{c_2}{c_1}f_2(x)\]

Conversely, if we suppose

\[f_1(x) = c_2f_2(x)\]

Then

\[(-1)f_{1}(x)+c_{2}f_{2}(x)=0, \quad \forall x \in I\]

So that the functions are linearly dependent because \(c_1 = -1\).

Conclusion

• \(f_1(x)\) and \(f_2(x)\) are linearly dependent on an interval⁵ \(I\) if and only if one is the constant multiple of other.
• \(f_1(x)\) and \(f_2(x)\) are linearly independent on an interval⁵ \(I\) if neither is the constant multiple of other.
• In general, a set⁶ of \(n\) functions¹ is linearly dependent if at least one of them can be expressed as a linear combination of the remaining.

Example

Consider the functions¹

\[f_{1}(x)=1+x, \quad \forall x \in (-\infty, \infty)\]

\[f_{2}(x)=x, \quad \forall x \in (-\infty, \infty)\]

\[f_{3}(x)=x^{2}, \quad \forall x \in (-\infty, \infty)\]

Then

\[c_1f_1(x)+c_2f_2(x)+c_3f_3(x)=0\]

means that

\[c_1(1+x)+c_2x+c_3x^2 = 0\]

or

\[c_1+(c_1+c_2)x+c_3x^2 = 0\]

Equating the coefficients of \(x\) and \(x^2\) constant terms, we obtain

\[c_1=0=c_3\]

\[c_1+c_2=0\]

Therefore

\[c_1=c_2=c_3=0\]

This shows that they all are linearly dependent.

Wronskian

Suppose the functions¹ \(f_1(x), f_2(x), \ldots, f_n(x)\) possess at least \(n - 1\) derivatives then the following determinant is called wronskian of functions.¹

\[ \begin{vmatrix} f_1 & f_2 & \cdots & f_n \\ f_1' & f_2' & \cdots & f_n' \\ \vdots & \vdots & \ddots & \vdots \\ f_1^{(n-1)} & f_2^{(n-1)} & \cdots & f_n^{(n-1)} \end{vmatrix} \]

and is denoted by

\[W(f_1(x), f_2(x), \ldots, f_n(x))\]

Theorem

The functions¹ \(f_1(x), f_2(x), \ldots, f_n(x)\) are linearly independent on the interval⁵ \(I\) if for at least one point on \(I\) such that

\[W(f_1(x), f_2(x), \ldots, f_n(x)) \ne 0\]

Also if

\[W(f_1(x), f_2(x), \ldots, f_n(x)) = 0\]

then they are linearly dependent but the converse of this is not true.

Example

The following functions¹ are linearly dependent

\[f_{1}(x)=\sin^{2}x\]

\[f_{2}(x)=1-\cos2x\]

because

\[\sin^{2}x=\frac{1}{2}(1-\cos2x)\]

We observe that for all \(x \in (-\infty, \infty)\)

\[ W(f_1(x), f_2(x)) = \begin{vmatrix} \sin^2(x) & 1 - \cos(2x) \\ 2 \sin(x) \cos(x) & 2 \sin(2x) \end{vmatrix} \]

\[=2\sin^{2}x\sin2x-2\sin x\cos x+2\sin x\cos x\cos2x\]

\[=\sin2x[2\sin^{2}x-1+\cos2x]\]

\[=\sin2x[2\sin^{2}x-1+\cos^{2}x-\sin^{2}x]\]

\[=\sin2x[\sin^{2}x+\cos^{2}x-1]\]

\[=0\]

References

Read more about notations and symbols.

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1. Read more about functions. ↩↩↩↩↩↩↩↩↩↩↩↩↩↩↩↩↩↩↩

2. Read more about slopes. ↩

3. Read more about continuous. ↩↩

4. Read more about linear differential equations. ↩

5. Read more about intervals. ↩↩↩↩↩↩

6. Read more about sets. ↩↩↩