First-Order Differential Equations

## Concept of Order

The order of a differential equation is the order of the highest derivative that appears in the equation.
For example,
y' = cos x                          (First order differential equation),
y'' + 4y= 0                       (Second order differential equation)
x2 y'''y' + 2y'' = x2 y2         (Third order differential equation)
The first-order differential equations contains only y' and may contain y and given functions of x . Hence we can write
F (x, y, y') = 0 or y' = f (x, y)        ………(1)

## Concept of Solution

A solution of a given first-order differential equation (1) on some open interval a < x< b is a function y =h (x) that has derivative y' = h' (x) and satisfies (1) for all x in that interval. Thus equation (1) becomes an identity if we replace the unknown function y by h (x) and y' by h' (x) .
Example: Verify that y = x 2 is a solution of xy' = 2y for all x .
Substitute y = x2 and  y' = 2x into xy' = x x 2x = 2x2 = 2y , an identity in x .

Implicit Solution

Sometimes a solution of differential equation will appear as an implicit function, i.e.
H(x,y) = 0,
and is called an implicit solution, in contrast to an explicit solution y = h (x) .
Example: The function y of x implicitly given by x2+y- 1 = 0 (y > 0), represents a semicircle of unit radius in the upper-half plane. This function is an implicit solution of the differential equation yy' = -x on the interval -1< x < 1.
General and Particular Solution
Consider the differential equation y' = cos x .
Its solution will be y = sin x + c where c is an arbitrary constant. Such a function involving an arbitrary constant is called a general solution of a first order differential equation.
If we choose specific c(c = 2 or 0 or -5/3, etc), we obtain what is called a particular solution of that equation.
Thus y = sin x + c is a general solution of y' = cos x , and y = sin x , y = sin x - 2 , y = sinx + 0.75 , etc. are particular solutions.
Singular Solution
A differential equation may sometimes have an additional solution that can not be obtained from the general solution and is then called a singular solution.
For example, y'2 - xy' + y = 0 has the general solution y = cx- c2. Substitution also shows that the is also a solution. This is a singular solution because we cannot obtain it from y = cx- c2 by choosing a suitable c .
Initial Value Problems
A differential equation together with an initial condition is called an initial value problem. It is of the form
y' = f(x, y), y(x0) = y0
The initial condition y(x0) = yis used to determine a value of c in the general solution.

### Separable Differential Equations

If a differential equation can be written in the form
f(y)dy = ∅(x)dx
We say that variables are separable, y on the left hand side and x on the right hand side. We get the solution by integrating both sides.

Example 1: Solve the differential equation 9 yy' + 4 x = 0 .

By separating variables we have
9 ydy =-4 xdx
Integrating both sides, we get

The solution represents a family of ellipses.

Example 2: Solve the differential equation y' = 1 + y2 .

By separating variables we have

Integrating both sides, we get
tan-1 y = x + c ⇒ y = tan (x + c)

Example 3: Solve the differential equation (yx2 + y) dy + (xy2 + x) dx = 0 .

Example 4: Solve the differential equation

By separating variables we have

Integrating both sides, we get

Example 5: Solve the differential equation y' = -2 xy , with y (0) = 1.

By separating variables we have

Integrating both sides, we get

Setting  when y > 0 , and when y < 0 , and admitting also c = 0 (which gives the solution y = 0 ), we get the general solution
y = ce -x2
∵ y(0) = 1⇒ c = 1 ⇒ y ⇒ e -x2

Example 6: Solve the differential equation

### Homogeneous Equation (Reduction to Separable Form)

A differential equation of the form

is called a homogeneous equation if each term of f (x, y) and ∅(x, y) is of the same degree.

Example 7: Find the solution of the differential equation 2 xyy' = y2 - x2 .

This general solution represents the family of circles with centers on the x -axis and all passing through origin.

Example 8: Find the solution of the differential equation

### Equations Reducible to Homogeneous Form

The equations of the form can be reduced to homogeneous form by the substitutions x = X + h, y = Y + k ( h,k being constants)

Case of failure:
Now put ax + by = z and apply the method of separation of variables.

Example 9:

Put x = X + h, y = Y + k (h, k being constants)
The given equation reduces to
Now choose h, k so that

On integrating we have

Put X = x - 1,Y = y - 1 ⇒ x+ y - 2 = a(x - y)3.

Example 10:

Put x + 2y = z ⇒

### Exact Differential Equation

If a function u (x, y) has continuous partial derivatives, its differential is

From this it follows that if u (x, y) = c = constant , then du = 0.
For example, if u = x + x2 y3 = c , then

A differential equation that we can solve by going backward.
A first-order differential equation of the form
M (x, y) dx + N (x, y) dy = 0                      …….(1)
is called  an exact differential equation if differentia form M (x,y) dx + N(x,y) dy is exact, that is, this form is the differential

of some function u (x, y).
Then the differential equation (1) can be written
du = 0 .
By integrating we obtain general solution of (1) in the form
u (x, y) = c    …….(3)
Comparing (1) and (2), we see that (1) is an exact differential if there is some function u (x, y) such that

Thus
This condition is not only necessary but also sufficient for (1) to be an exact differential equation.

In this integration, y is to be regarded as a constant, and k (y) plays the role of a “constant” of integration. To determine k (y) , use and find then integrate it to get k .
Similarly,
In this integration, x is to be regarded as a constant, and l (x) plays the role of a “constant” of integration. To determine and find then integrate it to get k .

Example 11: Solve (x3 + 3xy2) dx + (3x2y + y3) dy = 0

1st Step

This is an exact differential equation.
2nd Step

Example 12: Solve (sin x cosh y) dx - (cos x sinh y)dy = 0 ,   y (0)  = 3 .

1st Step
M = (sin x coshy) , N = (cos x sinh y) . Thus
This is an exact differential equation.
2nd Step

Thus u = - cosx cosh y + c' ⇒ u= cosx cosh y = c
∵ y (0) = 3 ⇒ cos 0 cosh 3 = 10.07 = c
⇒ cos x coshy = 10.07

### Equations Reducible to the Exact Form

Consider the equation
- ydx + xdy = 0 .

Hence equation is not exact. But if we multiply it by  we get an exact equation,

All we have done was the multiplication of a given nonexact equation, say
P (x, y) dx + Q (x, y) dy= 0
by a function F (x, y) . The result was an equation
FPdx + FQdy = 0
That is exact. The function F = F (x, y) is then called an integrating factor of
P (x, y) dx + Q (x, y) dy= 0

How to Find Integrating Factors
For FPdx + FQdy = 0 to be exact

In the general case, this would be complicated and useless.
(a) For simplification let F = F (x) , Fy = 0, / Fx = F' = dF/dx

Integrating Factor F (x) =
(b) If F = F (y), Fx = 0, Fy = F' = dF/dy

Integrating Factor

Example 13: Solve 2sin (y2) dx + xy cos (y2) = dy = 0,

1st step: Check for exactness. We have
P = 2sin (y2)  and  Q = xy cos (y2)
The equation is not exact because
Py = 4ycos (y2) ≠ Qx = y cos (y2)
2nd step: Integrating factor

Thus the integrating factor is

Multiplying the given equation by x3, we get
2x3sin (y2) dx + x4 y cos(y2) dy = 0
This equation is exact because

3rd Step: General Solution

Hence k ' (y) = 0 and k = const. this give the general solution

4th step: Particular solution
Substituting the initial condition into u (x, y), we have

Hence the desired particular solution is

### Linear Differential Equations

A first-order differential equation is said to be linear if it can be written as

where p and r are function of x (but not y) or constant.
If r(x) = 0 , the equation is said to be homogeneous; i.e.

If r (x) ≠ 0 , the equation is said to be nonhomogeneous; i.e.

Compare with Pdx + Qdy = 0 , thus P = (py - r), Q = 1

Integrating Factor F (x) =
Multiplying

Example 14: Solve the linear differential equation y' - y = e2x .

Here  p = -1, r = e2x ,
Thus general solution

In simpler cases, such as the present, we may not need the general formula but may wish to proceed directly; multiplying the given equation by eh = e-x . This gives

Example 15: Solve y' + 2y = ex (3sin 2x + 2 cos 2 x)

Here p = 2, r = ex (3 sin 2 x + 2 cos 2 x)
Thus general solution

Example 16: Solve the initial value problem

y' + y tan x = sin 2x ,   y (0) = 1

Here p = tan x,r = sin 2 x = 2 sin x cos x and
The general solution is

⇒ sec x x y =

From this and the initial condition 1 = c - 1- 2.12 ; thus c = 3 and the solution of our initial value problem is y = 3cosx- 2 cos2 x.

### Equation Reducible to Linear Form

A differential equation of the form
where p and q are function of x (but not y) or constant can be reduced to the linear form on dividing by yn and substituting

which is a linear differential equation.

Example 17: Solve y' - Ay= - By2

Divide above equation by y2,

This gives the general solution

The document First-Order Differential Equations | Mathematical Methods - Physics is a part of the Physics Course Mathematical Methods.
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## Mathematical Methods

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## FAQs on First-Order Differential Equations - Mathematical Methods - Physics

 1. What is the concept of order in first-order differential equations?
Ans. In first-order differential equations, the concept of order refers to the highest derivative appearing in the equation. For example, if the equation involves only the first derivative of the unknown function, it is called a first-order differential equation.
 2. What is the concept of a solution in first-order differential equations?
Ans. The concept of a solution in first-order differential equations refers to finding a function that satisfies the given equation. A solution is a function that, when differentiated, yields the expression present in the equation. It satisfies the equation for all possible values of the independent variable.
 3. What is the IIT JAM exam?
Ans. The IIT JAM (Joint Admission Test for M.Sc.) is an entrance exam conducted by the Indian Institutes of Technology (IITs) for admission to their various M.Sc., Joint M.Sc.-Ph.D., M.Sc.-Ph.D. dual degree, and other post-bachelor's degree programs. It is a national-level exam that tests the candidate's knowledge in various science subjects.
 4. What are some frequently asked questions in the IIT JAM exam related to the concept of first-order differential equations?
Ans. Some frequently asked questions in the IIT JAM exam related to first-order differential equations could include: - How to solve a first-order linear differential equation? - What is the concept of an integrating factor in solving first-order differential equations? - How can separation of variables be used to solve a first-order differential equation? - What are the initial conditions and how do they affect the solution of a first-order differential equation? - Can a first-order differential equation have multiple solutions?
 5. Can you provide some tips for preparing for the IIT JAM exam and mastering the concept of first-order differential equations?
Ans. Some tips for preparing for the IIT JAM exam and mastering the concept of first-order differential equations include: - Understand the basic concepts and properties of first-order differential equations. - Practice solving a variety of problems to improve your problem-solving skills. - Familiarize yourself with different solution techniques such as separation of variables and integrating factors. - Review and revise the important formulas and concepts regularly. - Solve previous years' question papers and take mock tests to assess your preparation level and time management skills.

## Mathematical Methods

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