Unit1 - Subjective Questions

Definition:
A differential equation of the form is said to be exact if it can be derived from its primitive (general solution) directly by differentiation, without any subsequent multiplication, elimination, or reduction.

Necessary and Sufficient Condition:
The necessary and sufficient condition for the differential equation to be exact is:

Where:

• and are functions of and .
• The partial derivatives represent the rate of change of with respect to (keeping constant) and with respect to (keeping constant).

Step 1: Identify M and N
Comparing with :

Step 2: Check for Exactness


Since , the equation is exact.

Step 3: Apply the Solution Formula
The solution is given by:

Calculation:

2. Terms of free from : Only .

Final Solution:

Or multiplying by 3:

If a differential equation is homogeneous (i.e., both and are homogeneous functions of the same degree) and is not exact (), an Integrating Factor can be found using the following rule:

Rule:
If , then the integrating factor is:

Procedure:

1. Verify homogeneity of and .
2. Calculate .
3. Ensure the result is not zero.
4. Multiply the original differential equation by .
5. The resulting equation will be exact and can be solved using the standard method.

Here and .
Both and are homogeneous of degree 3.

Calculate Mx + Ny:

Integrating Factor:

Multiply Equation by I.F.:

Solve Exact Equation:

Terms in new free from :

General Solution:

For a differential equation of the form:

Where and .

Rule:
If , then the integrating factor is given by:

Multiplying the given equation by this factor reduces it to an exact differential equation.

, .

Difference: .

Check Rule (): .

This is a function of only. So, .

Multiply equation by y:

Solve:

Terms in free from :

Solution:

Definition:
A differential equation of the first order and degree is of the form:

where and are functions of and .

Standard Methods of Solution:

1. Equations solvable for p: The equation can be factorized into linear factors of .
2. Equations solvable for y: The equation can be expressed as .
3. Equations solvable for x: The equation can be expressed as .

The given equation is quadratic in ().

Step 1: Factorize

Step 2: Solve individual components
Case 1:
Integrating:

Case 2:
Integrating:

Step 3: General Solution
The general solution is the product of the individual solutions (using a single constant ):

If the equation can be written in the form , follow these steps:

1. Differentiate with respect to x:
   Differentiate the equation w.r.t . Remember that .
   

2. Solve for p:
   The resulting equation will be a differential equation in variables and . Solve this to find a relation between , and an arbitrary constant .
   Let this solution be .

3. Eliminate p:
   Eliminate between the original equation and the solution to get the general solution in terms of and .

   Note: If elimination is difficult, express both and in terms of parameter to get the solution in parametric form.

Given ... (1)

Differentiate w.r.t x:

Separate Variables:

Integrate:
... (2)

General Solution:
Equations (1) and (2) together form the parametric solution:

(Eliminating is algebraically complex here, so parametric form is acceptable).

If the equation can be written in the form , follow these steps:

1. Differentiate with respect to y:
   Differentiate the equation w.r.t . Remember that .
   

2. Solve for p:
   The resulting equation is a first-order differential equation in variables and . Solve it to get a relation .

3. Eliminate p:
   Eliminate between the original equation and to obtain the general solution.
   Alternatively, present the solution in parametric form ( and in terms of ).

Given (Solvable for x).

Differentiate w.r.t y:

Since :

Multiply by :

Separate Variables:

Integrate:

General Solution:
Substitute back into the original equation for :

The solution is given parametrically by:

Definition:
Clairaut's equation is a differential equation of the first order and higher degree of the form:

where .

Derivation of General Solution:

1. Differentiate the equation w.r.t :
   
   Since :
   
   

2. This gives two factors:

   • Factor 1:
   • Factor 2:

3. For General Solution, take . Integrating gives (constant).

4. Substitute into the original equation:
   

This is the general solution.

This equation is in the standard Clairaut’s form:

where .

The general solution for Clairaut's equation is obtained simply by replacing with an arbitrary constant .

Solution:

Step 1: Rearrange to identify form
Divide by :

Step 2: Identify Form
This is Clairaut's Equation of the form , where .

Step 3: Write General Solution
Replace with constant :

Simplifying:

General Solution:
The equation is in Clairaut's form ().
Replace with :

Singular Solution:
The singular solution is the envelope of the family of lines given by the general solution. We differentiate the general solution partially w.r.t and equate to zero (c-discriminant method).

2. Differentiate w.r.t :
   
   

3. Substitute back into the general solution:
   
   
   
   

is the singular solution.

Substitution:
Let and .
Then and .
Let .

Substitute into Equation:

Divide by :

Multiply by :

Substitute back and :

Solve:
This is Clairaut's form in variables .
General Solution: .

Restore variables:

Rearrange as a quadratic in :

Solve for using quadratic formula:

Integrate:

This yields two solutions based on the sign chosen.

Consider the equation . Let be the integrating factor dependent only on .
Multiplying by , the equation becomes exact.

Condition for exactness:

Since is a function of only, .
So,

Rearranging:

If the RHS is a function of only, say , then:

, .


Wait, checking values: , .

Correction: The equation is already exact.

Solve as Exact:

Terms in free from : .

Solution: