Linear Ordinary Differential Equations of First and Second Order - 1 | Physics for IIT JAM, UGC - NET, CSIR NET PDF Download

We consider the equation

The function f (t, y) is a linear function in y, i.e, we can write
f (t, y) = −p(t)y + g(t).
So we will study the equation
y′ + p(t)y = g(t).                                              (A)
We introduce the method of integrating factors (due to Leibniz): We multiply equation (A) by a function µ(t) on both sides
µ(t)y′ + µ(t)p(t)y = µ(t)g(t)
The function µ is chosen such that the equation is integrable, meaning the LHS (Left Hand Side) is the derivative of something. In particular, we require:
µ(t)y′ + µ(t)p(t)y = (µ(t)y)′ , ⇒ µ(t)y′ + µ(t)p(t)y = µ(t)y′ + µ′(t)y
which requires

Integrating both sides

which gives a formula to compute µ

Therefore, this µ is called the integrating factor.

Note that µ is not unique. In fact, adding an integration constant, we will get a different µ. But we don’t need to be bothered, since any such µ will work. We can simply choose one that is convenient.

Putting back into equation (A), we get

which give the formula for the solution

Example 1. Solve y′ + ay = b (a = 0).
Answer. For b = 0 the homogeneous first-order linear differential equation with constant coefficients is available.
y' + ay = 0
Solution of the homogeneous linear differential equation of first order with constant coefficients:
y' = −ay
Transformation of equation
y'/y = −a
Division by y
(ln y)' = −a
Applying the chain rule
ln y = −a∫dx = −ax + C
Integration yh = Ce^−ax 
General solution of the homogeneous equation with undetermined constants C
Variation of the constants: The solution of the inhomogeneous differential equations can be obtained from the homogeneous one. Generally the solution of the inhomogeneous equation is given by the solution of the homogeneous equation plus a special solution of the inhomogeneous equation. The special solution can be obtained by the method of variation of constants. Here the constant C of the homogeneous solution is assumed as a function of x and the homogeneous solution is inserted into the inhomogeneous equation. C(x) is then determined so that the equation is fulfilled.
y′_h = C'e^−ax − aCe^−ax
Derivation of the homogeneous solution with C as a function of x
C'e^−ax − aCe^−ax + aCe^−ax = b
Insertion into the inhomogeneous equation
C' = be^ax 
By rearranging we obtain an equation for the determination of C
C = b/ae^ax Integration gives C(x)
ys = b/a Insertion of C(x) in yh provides a special solution y_s 
y = ys + yh = b/a+Ce^−ax 
This is the general solution of the inhomogeneous differential equation with constant coefficients

Example 2. Solve y′ + y = e^2t.
Answer. We have p(t) = 1 and g(t) = e^2t. So

and

Example 3. Solve

Answer. First, let’s rewrite the equation into the normal form

so

Then

Then

By the IC y(0) = 1:

Example 4. Solve ty′ − y = t² e^−t , (t > 0).
Answer. Rewrite it into normal form

so
p(t) = −1/t, g(t) = te^−t.
We have

and

Example 5. Solve  with y(0) = a, and discussion the behavior of y as t → ∞, as one chooses different initial value a.
Answer. Let’s solve it first. We have

so

Plug in the IC to find c

so

To see the behavior of the solution, we see that it contains two terms. The first term e^−t goes to 0 as t grows. The second term e^t/3 goes to ∞ as t grows, but the constant

On the other hand, as t → −∞, the term e^−t will blow up to −∞, and will dominate.
Therefore, y → −∞ as t → −∞ for any values of a.
See plot below:

Example 6. Solve ty′ + 2y = 4t² , y(1) = 2.
Answer. Rewrite the equation first

So p(t) = 2/t and g(t) = 4t. We have

and

By IC y (1) = 2
y(1) = 1 + c = 2, c = 1
we get the solution:

Note the condition t > 0 comes from the fact that the initial condition is given at t = 1, and we require t ≠ 0.
In the graph below we plot several solutions in the t − y plan, depending on initial data.
The one for our solution is plotted with dashed line where the initial point is marked with a
‘×’.

Separable Equations 

We study first order equations that can be written as

where M (x) and N (y) are suitable functions of x and y only. Then we have

and we get implicitly defined solutions of y(x).

Example 1. Consider
Answer. We can separate the variables:

If one has IC as y(π) = 2, then

so the solution y(x) is implicitly given as

Example 2. Find the solution in explicit form for the equation

Answer. Separate the variables

Set in the IC y(0) = −1, i.e., y = −1 when x = 0, we get

In explicitly form, one has two choices:

To determine which sign is the correct one, we check again by the initial condition:
 must have y(0) = −1.
We see we must choose the ‘-’ sign. The solution in explicitly form is:

On which interval will this solution be defined?

We can also argue that when x = −2, we have y = 1. At this point |dy/dx| → ∞, therefore solution can not be defined at this point.
The plot of the solution is given below, where the initial data is marked with ‘x’. We also include the solution with the ‘+’ sign, using dotted line.

Example 3. Solve y′ = 3x² + 3x² y², y(0) = 0, and find the interval where the solution is defined.
Answer. Let’s first separate the variables.

Set in the IC:
arctan 0 = 0 + c, ⇒ c = 0
we get the solution
arctan y = x³ , ⇒ y = tan(x³).
Since the initial data is given at x = 0, i.e., x³ = 0, and tan is defined on the interval  
We have

Example 4. Solve

and identify the interval where solution is valid.
Answer. Separate the variables

Set in the IC: x = 0, y = 1, we get
1 − 3 = c, ⇒ c = −2,
Then,
y³ − 3y² = x³ − x − 2.
Note that solution is given in implicitly form.
To find the valid interval of this solution, we note that y′ is not defined if 3y² − 6y = 0, i.e., when y = 0 or y = 2. These are the two so-called “bad points” where you can not define the solution. To find the corresponding values of x, we use the solution expression:
y = 0: x³+x−² = 0,
⇒ (x² + x + 2)(x − 1) = 0, ⇒ x = 1
and

(Note that we used the facts x² + x + 2 = 0 and x² − x + 2 = 0 for all x.) Draw the real line and work on it as following:

Therefore the interval is −1 < x < 1.

Differences between linear and nonlinear equations

We will take this chapter before the modeling .
For a linear equation
y′ + p(t)y = g(t), y(t0) = y₀,
we have the following existence and uniqueness theorem.
Theorem: If p(t) and g(t) are continuous and bounded on an open interval containing t₀, then it has an unique solution on that interval.

Example 1. Find the largest interval where the solution can be defined for the following problems.
(A). ty′ + y = t³, y(−1) = 3.

Answer. The equation is same as (A), so t = 0. t0 = 1, the interval is t > 0.
(C). (t − 3)y′ + (ln t)y = 2t, y(1) = 2
Answer. Rewrite:  and t > 0 for the ln function. Since t0 = 1, the interval is then 0 < t < 3.
(D). y′ + (tan t)y = sin t, y(π) = 100.
Answer. Since t0 = π, and for tan t to be defined we must have  
So the interval is
For non-linear equation
y′ = f (t, y), y(t₀) = y₀,
we have the following theorem:
Theorem. If   are continuous and bounded on an rectangle (α < t < β , a < y < b) containing (t0, y0 ), then there exists an open interval around t₀, contained in (α, β ), where the solution exists and is unique.
We note that the statement of this theorem is not as strong as the one for linear equation.
Below we give two counter examples:
Example 1. Loss of uniqueness. Consider
 y(−2) = 0.
Answer. We first note that at y = 0, which is the initial value of y, we have y′ = f (t, y) → ∞. So the conditions of the Theorem are not satisfied, and we expect something to go wrong.
Solve the equation as an separable equation, we get

and by IC we get c = (−2)2 + 0 = 4, so y2 + t2 = 4. In the y − t plan, this is the equation for a circle, centered at the origin, with radius 2. The initial condition is given at t₀ = −2, y₀ = 0, where the tangent line is vertical (i.e., with infinite slope). We have two solutions: y =     and  y =

We lose uniqueness of solutions.

Example 2. Blow-up of solution. Consider a simple non-linear equation: 
y′ = y², y(0) = 1.
Answer. Note that f (t, y) = y², which is defined for all t and y. But, due to the non-linearity of f , solution can not be defined for all t.
This equation can be easily solved as a separable equation.

By IC y(0) = 1, we get 1 = −1/(0 + c), and so c = −1, and
We see that the solution blows up as t → 1, and can not be defined beyond that point.
This kind of blow-up phenomenon is well-known for non-linear equations.