Second order linear differential equation with variable coefficients examples pdf

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

\(\begin{array}{l}\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}+ P \frac{\mathrm{d} y}{\mathrm{d} x}+Qy =R\end{array} \)

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 ao, a1 and b such that ao(t), a1(t) and b(t) ∈ R ∀ t ∈ R, the differential equation of the form

y” + a1(t) y’ + ao(t) y = b(t)

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

Here,

\(\begin{array}{l}y”= \frac{\mathrm{d^{2}}y }{\mathrm{d} x^{2}}\;\:and \;\: y’= \frac{\mathrm{d} y}{\mathrm{d} x}\end{array} \)

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

y” + 2t y’ + loge t y = e3t is a second-order differential equation 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 tmy(n)(t), its Laplace transform is (– 1)m dm/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

\(\begin{array}{l}\Rightarrow -\frac{d}{ds}L\left\{ y”\right\} + L\left\{ y’\right\}+2\frac{d}{ds}L\left\{y’\right\}-2L\left\{y \right\}=0\end{array} \)

\(\begin{array}{l}\Rightarrow -\frac{d}{ds}[s^{2}L\left\{ y\right\}-sy(0)-y'(0)] +[sL\left\{ y\right\}-y(0)]+2\frac{d}{ds}[sL\left\{ y\right\}-y(0)]-2L\left\{ y\right\}=0\end{array} \)

\(\begin{array}{l}\Rightarrow -\frac{d}{ds}(s^{2}z-s-2)+(sz-1)+2\frac{d}{ds}(sz-1)-2z = 0 \;\;\: where\;\;\;z=L\left\{ y\right\}\end{array} \)

\(\begin{array}{l}\Rightarrow -(s^{2}-2s)\frac{dz}{ds}-sz = 0\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{dz}{z}+\frac{ds}{s-2}=0\end{array} \)

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 e2t

⇒ y = C1 e2t

But y(0) = 1, therefore C1 = 1

Hence, the solution of the given differential equation is y = e2t.

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

Wy1y2(t) = y1(t) y2‘(t) – y2(t) y1‘(t) ∀ t ∈ R

Alternatively,

\(\begin{array}{l}W_{y_{1}y_{2}}(t)=\begin{vmatrix}y_{1}(t) &y_{2}(t) \\ y_{1}'(t) &y_{2}'(t) \\\end{vmatrix}\end{array} \)

Fundamental Solution

If y1 and y2 are two solutions of the differential equation y” + a1(t) y’ + ao(t) y = 0, then y1 and y2 are called fundamental solution if and only if y1 and y2 are linearly independent, that is, Wy1y2 ≠ 0.

General Solution

If y1 and y2 are two fundamental solution of the differential equation y” + a1(t) y’ + ao(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,

\(\begin{array}{l}I.F. = e^{\int (s + 2/s)ds}= e^{\frac{s^{2}}{2}+2ln(s)}= s^{2}e^{\frac{s^{2}}{2}}\end{array} \)

The solution of (1) is

\(\begin{array}{l}dz. s^{2}e^{s^{2}/2}= \int \left ( 1 + 2/s + 1/s^{2} \right )s^{2}e^{s^{2}/2}ds + C_{1}\end{array} \)

\(\begin{array}{l}=\int (2u+1)e^{u}. \frac{du}{\sqrt{2u}} + 2\int \sqrt{2u}. e^{u}. \frac{du}{\sqrt{2u}}+ C_{1}\end{array} \)

Putting s2/2 = u so that s ds = du

Or,

\(\begin{array}{l}ds = \frac{du}{\sqrt{2u}} = \int \sqrt{2u}e^{u}du + \int \frac{e^{4}}{\sqrt{2u}}du + 2 \int e^{u}du+ C_{1}\end{array} \)

\(\begin{array}{l}=\sqrt{2u}e^{u}+2e^{u}+ C_{1}= se^{s^{2}/2}+ 2e^{s^{2}/2}+ C_{1}\end{array} \)

Or,

\(\begin{array}{l}z = L\left\{ y\right\} = \frac{C_{1}}{s^{2}}e^{-s^{2}/2}+1/s + 1/s^{2} = \frac{2+C_{1}}{{s^{2}}} -\frac{C_{1}}{2}+ \frac{C_{1}}{8}s^{2}-…+\frac{1}{s}\end{array} \)

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 Wy1y2 ≠ 0

y1(t) = √t and y1‘(t) = ½(t – ½ )

y2 = 1/t and y2‘(t) = – t – 2

Now, Wy1y2 = t1/2 (– 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.

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