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

First-order differential equations involve the first derivative of an unknown function y (t). They are of the form:
y′ = f(t, y),
where y′ = dy/dt, and f (t, y) is a given function. These notes cover two main types: linear and separable equations, with a focus on solution methods and exam-relevant techniques.

Linear First-Order Differential Equations

A linear first-order differential equation has the form:
y′ + p(t)y = g(t),
where p (t) and g (t) are functions of t. The solution method is the integrating factor technique.

Method of Integrating Factors

Multiply both sides of the equation by an integrating factor μ (t):  μ(t)y′ + μ(t)p(t)y = μ(t)g(t).
Choose μ (t) such that the left-hand side is the derivative of a product:
By the product rule, d/dt [μ (t) y] = μ (t) y ′ + μ ′ (t) y . Equate coefficients of y:
μ′(t)y = μ(t)p(t)y ⟹ μ′(t) = μ(t)p(t).
This simplifies to:
Thus, the integrating factor is:
The constant of integration can be omitted, as any μ (t) of this form works. The equation becomes:

Integrate both sides:

Solve for y:

Examples

Example 1: Solve y ′ + ay = b , where a ≠ 0 , with no initial condition

Here, p (t) = a , g (t) = b. The integrating factor is:

Multiply through:

The left-hand side is:

Integrate:

Solve for y :

For the homogeneous case (b = 0 ):

The general solution combines the homogeneous and particular solutions, confirming the result.

Example 2: Solve y ′ + y = e^2t

Here, p (t) = 1 , g (t) = e^2t . The integrating factor is:
Multiply through:
The left-hand side is:

Integrate:

Solve for y:

Verify by substituting back to ensure correctness (a good exam practice).

Example 3: Solve ty ′ + y = t³, y ( − 1 ) = 3

Rewrite in standard form (t ≠ 0):

Here, p(t) = 1/t, g (t) = t². The integrating factor is:

Since y (− 1) = 3 implies t < 0 , use μ (t) = − t for t < 0 . Multiply through:

The left-hand side is:

Integrate:

Solve for y:

Apply the initial condition y (− 1) = 3:

Thus:
The solution is valid for t < 0 , as t = 0 makes the denominator undefined.

Example 4: Solve t y ′ − y = t²e ^{− t}, t > 0

Rewrite in standard form:

Here, p(t) = -1/t, g (t) = te^−t. The integrating factor is:
Multiply through:
The left-hand side is:

Integrate:
Solve for y :

No initial condition is provided, so this is the general solution for t > 0.

Separable Equations

A separable equation is of the form:

where M (x) and N (y) depend only on x and y , respectively. 
Rewrite as: N(y)dy = M(x)dx.
Integrate both sides to find the solution, often implicitly:

Examples

Example 1: Solve y ′ = − 2x/y, y (π) = 2

Rewrite: y dy = −2x dx.
Integrate:

Apply the initial condition y (π) = 2:
4 = −2π² + K ⟹ K = 4 + 2π².
Thus,
y2 = −2x² + 4 + 2π².
Solve explicitly:
Since y (π) = 2, take the positive root. The solution is valid where − 2x² + 4 + 2π² > 0.

Example 2: Solve y ′ = 3x²(1 + y²) , y (0) = 0

Rewrite:Integrate:
Apply the initial condition y (0) = 0:
arctan 0 = 0 + C ⟹ C = 0.
Thus:
arctan y = x³ ⟹ y = tan(x³).
The solution is defined where

Example 3: Solve

Rewrite: y²(3 − y)dy = x²dx.
Integrate:Left-hand side:
Right-hand side:

Thus:

Apply the initial condition y (0) = 1:Thus:
To find the valid interval, note that y ′ is undefined at y = 0 or y = 3 . Check for x values:
At y = 0:

At y = 3:

Since y (0) = 1 , the solution is valid between these points: − (9/4)^1/3 < x < 18^1/3.

Existence and Uniqueness

For a linear equation y ′ + p (t) y = g (t), y (t₀) = y0:

• If p (t) and g (t) are continuous on an open interval containing t₀, the solution exists and is unique on that interval.

For a nonlinear equation y′ = f(t, y), y (t₀) = y0:

• If f and ∂f/∂y are continuous in a rectangle around (t₀, y₀), a unique solution exists in some interval around t₀.

Examples

Example 1: Largest interval for ty′ + y = t³, y (− 1) = 3

Standard form:

Here, p (t) = 1/t , g (t) = t², undefined at t = 0 . Since t₀ = − 1 , the solution is valid for t < 0.

Example 2: Largest interval for (t − 3) y ′+ (ln t) y = 2 t , y (1) = 2

Standard form:

Here,

undefined at t = 0 (due to ln t ) and t = 3. Since t₀ = 1 , the interval is 0 < t < 3.

Example 3: Blow-up for y ′ = y², y ( 0 ) = 1

Separate variables:

Apply y(0) = 1:

Thus:

The solution blows up at t = 1 , undefined beyond this point.