Class 12 chapter 9 (differential Equations) Class Notes

Differential Equations

Introduction

Differential equation plays an important role in the modern science.

In Physics, “Newton’s Law of cooling”- This states that the rate of change of the temperature of an object is directly proportional to the difference between its own temperature and the ambient temperature”.

\[\dfrac{{dT}}{{dt}} =  - K\left[ {T(t) - {T_a}(t)} \right]\]

This is differential equation

Solving a RC, RLC or RL circuit in physics involves differential equation.

\({V_{emf}} = {K_e}\dfrac{{d\theta }}{{dt}}\)   -- A differential Equation

Hence, an in-depth study of differential equations has assumed prime importance in all modern scientific investigations.

 

What is differential equation?

An equation that involves a relationship between a function and one or more derivatives is called a differential equation.

Example:  \(a\dfrac{{dy}}{{dx}} + x = y\)  A Derivative form \(\dfrac{{dy}}{{dx}} = \cos x\)

Above are the examples of differential equation since it involve a derivative form and the function in form of x and y.

But,
ax² + bx + c = 0 or \(x\sin x - x = 1\). These equations have variable x but do not include their derivatives hence these functions are not differential equation

Hyperbola equation: \(\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\), Hyperbola Equation is not a differential equation

  

Similarly y = sinx not a differential Equation  

Notation used: 

\(y' = \frac{{dy}}{{dx}}\) and \(y'' = \frac{{{d^2}y}}{{d{x^2}}}\), For n^th order \({y_n} = \frac{{{d^n}y}}{{d{x^n}}}\)

Order and Degree of Differential Equation

Order of differential Equation

The highest derivative in a differential equation is said to be a order of differential equation.

Example:

Now let’s find out order of the function : \(\dfrac{{{d^3}y}}{{d{x^3}}} + x{\left( {\dfrac{{dy}}{{dx}}} \right)^4} = 0\)

Order here is 3 because of the highest order of derivative i.e. d³y/dx³ 

Note: That there is difference in order and degree as shown in our next section.

Degree of differential Equation-
It is the highest power of the highest order derivative involved in the differential function.

Note:- The conditions to find the degree of differential equation is that the function should only be polynomial function if the differential equation contains log, exponential and trigonometric function of the derivative then degree is not defined i.e. the equation has to be polynomial function to define degree of a differential equation.

Example: \(\dfrac{{{d^3}y}}{{d{x^3}}} + x{\left( {\dfrac{{dy}}{{dx}}} \right)^4} = 0\)

Here the order is 3 and degree is 1 i.e the highest power of highest derivative. 

y''' + y' + y = 0          Here order is 3 and degree = 1.

\(\dfrac{{dy}}{{dx}} + \sin \left( {\dfrac{{dy}}{{dx}}} \right) = 0\)             – Here the order is 1 but degree is not defined.

Note: Order and degree (if defined) of a differential equation are always positive integers.

 

General and Particular Solution of Differential Equation

But for differential equation like y’-sinx=0. Solution of y is in function in terms of variables not in numbers.

We classify solution of differential equation into two parts-

1. General Solution
2. Particular Solution

Let us understand General solution of differential equation with an example.
 

Example: \(\frac{{dy}}{{dx}} - k\sin x = 0\)  .........{ A differential equation

Integrating both side \(\int {dy}  = \int {k\sin xdx} \)

We get solution as  y = - Kcosx + C 

This is called a “general solution” of a differential equation as we get the solutions in constant form of K and C (K and C can be any value from - ∞ to + ∞ )

Particular solution:
Now the solution we get is a General solution but if we assign some value to it by giving some conditions Let’s say we get the value of k and c be - 4 and 7 resp. then the equation will became y = 4cosx + 7    {This equation is a particular solution of a given differential equation}.

Formation of differential equation whose general solution is given

Now let’s learn how to form the differential equation of a given function. Please note that we need to eliminated the parameter like a,b  etc. to form a differential equation

Procedure to find a differential equation

1. If the given family F of curves depends on only one parameter then we can represent the equation of the form

F (x, y, a) = 0 ..(1)

Example: Parabola Equation y² = 4ax (as shown in fig.) this can be represented in an equation of the form  F(x, y, a) : y² = 4ax…..(1)

Now to find out differential equation we differentiate equation (1)
with respect to x, we get an equation involving  y^′, y, x, and a,

i.e.,   \(y\frac{{dy}}{{dx}} = 2a\) ……..(2)      

We get a differential equation of a family of a parabola by putting the value of parameter “a” from (2) in equation (1)

We get         \(y = 2x\frac{{dy}}{{dx}}\)   ………..this a differential equation.

 i.e  F(x,y,y’)=0

2. If two parameters are given say “a and b” in the function, then we represent the equation of the form F (x, y, a ,b) = 0

Example:  y = asin(x + b) this can be represented by an equation of the  form F(x, y, a, b) : y = asin(x + b)……….(1)

Now since it involve two parameter (a, b) we need to differentiate twice since it is not possible to eliminate two parameters a and b from the two equations only and so, we required a third equation.

\(\dfrac{{dy}}{{dx}} = a\cos (x + b)\)                       --(2)

\(\frac{{{d^2}y}}{{d{x^2}}} =  - a\sin (x + b)\)       --(3)

and since  we put the value in 3rd in the equation, we get \(\dfrac{{{d^2}y}}{{d{x^2}}} =  - y\)  or  \(\dfrac{{{d^2}y}}{{d{x^2}}} + y = 0\)

i.e F(x,y,y’,y”) = 0

Note:- Similarly we follow the same steps for 3 or more parameters if involved.

 

Problem:  Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.

Sol. The equation of the family of hyperbolas with the centre at origin and foci along the x-axis is: \(\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\)     ..(1)

           

Differentiating both sides of equation (1) with respect to x, we get:

\(\dfrac{{2x}}{{{a^2}}} + \dfrac{{2yy'}}{{{b^2}}} = 0\) or \(\dfrac{{x}}{{{a^2}}} + \dfrac{{yy'}}{{{b^2}}} = 0\)   ----(2)

Again Differentiating equation (2) with respect to x, we get: \(\dfrac{1}{{{a^2}}} - \dfrac{{y'y' + yy}}{{{b^2}}} = 0\) or \(\dfrac{1}{{{a^2}}} = \dfrac{{{{(y')}^2} + yy}}{{{b^2}}}\)     ....(3)

Now substituting value of \(\dfrac{1}{{{a^2}}}\) from equation in (3) in the equation (2) we get,

Solving the equation we get,

x(y’)²+xyy’’-yy’=0 or xyy’’+x(y’)²-yy’=0

This is a required differential equation of Hyperbola.

Differential Equations with Variables Separable

A first order-first degree differential equation is of the form

dy/dx = F(x,y)   --(1)

If F (x, y) can be expressed as a product g (x) h(y), where, g(x) is a function of x and h(y) is a function of y, then the differential equation (1) is said to be of “variable separable” type. Then differential equation (1) has the form             

dy/dx=  h(y).g(x)

d(y)/h(y) = g(x). dx   ... (3)   where h(y) ≠ 0

Integrating both sides of equation (3), we get

∫ (1/h(y)).dy =      ∫   g(x). dx                                                   =  ... (4)

Thus, (4) provides the solutions of given differential equation in the form

H(y) = G(x) + C

Here, H (y) and G (x) are the anti-derivatives of 1/h(y) and g(x) respectively and C is the arbitrary constant.

Problem: find the general solution of the following of question :  sec² x tan y dx + sec² y tan x dy = 0

Solution: Sec²y tanx dy= - sec² x tan y dx

Integrating both side 

Let tany=t and sec²ydy = dt

log(tany)= -log(tanx)+logC

tany.tanx=C

This is required general solution of differential equation.

 

Homogeneous Function

A function F(x, y) is said to be homogeneous function of degree n if

F(λx, λy) = λⁿ F(x, y) for any nonzero constant λ.

Eg:    F(x, y) = y² + 2xy    

F(λx, λy) = (λy)² + 2λxλy

F(λx, λy) = λ² (y² + 2xy )  

Thus, F(x, y) = y² + 2xy      is  a homogeneous function.

 

But,  F(x, y) = sin x + cos y is not a homogeneous function since it cannot be expressed as F(λx, λy) = λⁿ F(x, y)

Homogeneous Differential Equation

A differential equation of the form dy/dx = F (x, y) is said to be homogenous if F(x, y) is a homogenous function of degree zero

Example: Below Equation is Homogeneous Differential Equation

          

Since,

F(λx, λy) = λ⁰F(x,y)  

i.e  F(λx, λy) = F(x,y)

 

Method to solve a given homogeneous differential equation

     ………(1)

We make the substitution of    y = v . x  ……….(2)

Differentiating equation (2) with respect to x, we get

      ……..(3)

 

Substituting the value of  dy/dx  from equation (3) in equation (1), we get

   ………(4)

 

Arranging the variables,                                          ………(5)

 

Equation (5) gives general solution (primitive) of the differential equation (1) when we replace v by y/x

Note: If the homogeneous differential equation is in the form dy/dx= F(x, y) where, F (x, y) is homogenous function of degree zero, then we make substitution of x/y = v   i.e., x = vy and we proceed further to find the general solution as discussed above by writing dy/dx = (x,y) = g(y/x)

First order Linear differential equations

A differential equation of the from  dy/dx + Py = Q

where, P and Q are constants or functions of x only, is known as a first order linear differential equation.

Example:

Another form of first order linear differential equation can be dx/dy+ Px =Q

where, P and Q are constants or functions of y only, is known as a first order linear differential equation

Steps involved to solve first order linear differential equation:

(i) Write the given differential equation in the form dy/dx + Py = Q  ,  where P, Q are constants or functions of x only.

(ii) Find the Integrating Factor (I.F)

(iii) Write the solution of the given differential equation as

y (IF) =∫Q + IF. dx  + C

 

Problem:  Solve equation:     dy/dx + y/x = x² 

Solution. The given differential equation is in the form of dy/dx + Py = Q

(Here P=1/x and Q=x²)

 

This is the required general solution of the given differential equation.

 

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