Differential Equations | Engineering Mathematics - Civil Engineering (CE) PDF Download

Definition of Differential Equation

• A differential equation is a special kind of equation that includes terms involving the derivatives of one variable (known as the dependent variable) with respect to another variable (known as the independent variable).
• For instance, in the equation dy/dx = f(x), "x" is the independent variable, and "y" is the dependent variable.
• Differential equations can have either partial derivatives or ordinary derivatives. The derivative measures the rate of change, and the differential equation expresses a relationship between a quantity that changes continuously and the change in another quantity.
• There are various formulas and methods to solve these differential equations and find solutions for the derivatives.

Order of a differential equation

• Order of a differential equation is defined as the order of the highest order derivative of the dependent variable with respect to the independent variable involved in the given differential equation. Consider the following differential equations:
  dy / dx = e^x
  
• The first, second, and third equations involve the highest derivative of first, second, and third order respectively. Therefore, the order of these equations are 1, 2, and 3 respectively. 

Degree of a differential equation

• By the degree of a differential equation, when it is a polynomial equation in derivatives, we mean the highest power (positive integral index) of the highest order derivative involved in the given differential equation.

• To study the degree of a differential equation, the key point is that the differential equation must be a polynomial equation in derivatives, i.e., y′, y″, y″′ etc. Consider the following differential equations:
  
• We observe that first equation is a polynomial equation in y″′, y″ and y′ with degree 1,.
• Second equation  is a polynomial equation in y′ (not a polynomial in y though) with degree of 2. Degree of such differential equations can be defined.
• But third equation is not a polynomial equation in y′ and degree of such a differential equation can't be defined.

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

Example Find the order and degree, if defined, of each of the following differential equations:  
(i) dy/dx - cosx = 0
(ii) xy d²y/dx² + x(dy/dx)² - ydy/dx = 0
(ii) y''' + y² + e^y = 0
Solution 
(i) The highest order derivative present in the differential equation is dy / dx, so its order is one. It is a polynomial equation in y′ and the highest power raised to  dy / dx one, so its degree is one. 
(ii) The highest order derivative present in the given differential equation is d²y / dx², so its order is two. It is a polynomial equation in d²y / dx²and dy/dx and the highest power raised to d²y / dx² is one, so its degree is one. 
(iii) The highest order derivative present in the differential equation is y′′′, so its order is three. The given differential equation is not a polynomial equation in its derivatives and so its degree is not defined.

Separable differential equation

y' = A(x)B(y)

Example:

y' = y² e^-x   ⇒ y = 1/(e^-x+ c)
RC circuits: 

• Charging: 
• Discharging: IR = Q/C ⇒ Q = Q_oe^-1/RC

Linear Differential Equations

y'(x) + p(x)y(x) = q(x)

Integrating factor

I.F. = e^∫p(x)dx

The solution of the Differential Equation written above is given by:

y.(I.F.) = ∫q(x). (I.F. )dx + c

Example: Solve, y' = 3x²- y/x             y(1) = 5

Sol: y' + y/x = 3x²

I.F. = e^∫1/xdx = e^lnx = x

Therefore, y.x = ∫3x².x dx+ c

yx = 3x⁴/4 + c

Now    y(1) = 5

Hence, 5x1 = 3x1/4 + c

c = 5-3/4 = 17/4

There the solution of a given differential equation is y(x) = 3x⁴/4 + 17/4

Exact Differential Equations

(a) Potential function

For M(x, y) + N(x, y)y' = 0 we can find a ∅(x, y) such that d∅/dx = M and d∅ / dy = N;  is the potential function;  M(x, y) + N(x, y)y' is exact.

(b) Exact differential equation

A potential function exists; general solution: ∅(x, y) =
Example: 

Theorem: Test for exactness: 
Example:  x² = 3xy + (4xy + 2x)y'= 0
e^x sin y - 2x + (e^x cosy + 1)y' = 0

Integrating Factors

(a) Integrating factor: μ(x, y) ‡ 0 such that μM(x,y) + μN(x,y)y' = 0 is exact.

Example: y² - 6xy + (3xy - 6x²)y' = 0
(b) How to find integrating factor:
Example: x - xy - y'= 0
A. Separable equations and integrating factors: μ= 1/B
B. Linear equations and integrating factors: I.F. = e^∫p(x)dx

Homogeneous and Bernoulli Equations

Example: .
(b) Bernoulli equation:  y' = P(x)y = R(x)y^α; α= 0 ⇒ linear;α= 1
⇒ separable; otherwise, let v = y^1-α ⇒ linear
Example:

Higher Order Linear Ordinary Differential Equations

(i) Homogeneous Linear ODEs

An linear ordinary differential equation of order n is said to be homogeneous if it is of the form

a_n(x)y⁽ⁿ⁾ + a_{n - 1}(x)y^{(n - 1)} + .... + a₁(x)y' + a₀ (x)y = 0

where y'= dy/dx i.e., if all the terms are proportional to a derivative of y (or y itself) and there is no term that contains a function of x alone.
However, there is also another entirely different meaning for a first-order ordinary differential equation, such an equation is said to be homogeneous if it can be written in the form dy/dx = F(y/x).

(i) Homogeneous Linear ODEs with constant coefficients

1. : Substituting y= e^λx, we obtain the characteristic equation λⁿ + a_n-1λ^n-1 + ...+ a₁λ + a₀ = 0.
   (a) Distinct real roots: The general solution is
   Example: y''' - 2y'' - y' + 2y =0.
   Solution: y= c₁e^-x+ c₂e^x + c₃e^2x
   (b) Simple complex roots: , y₁ = e^px cos(qx), y2 = e^{px sin}(qx) .
   Example: y''' - y'' - 100y' + 100y =0.
   Solution: y = c₁e^x + c₂ cos10x + c3 sin10x.
   (c) Multiple real roots: If λ is a real root of order m, then m corresponding linearly independent solutions are: e^λx,xe^λx, x²e^λx,x^m-1e^λx .
   Example: y⁽⁵⁾ - 3y⁽⁴⁾ + 3y''' - y'' = 0.
   Solution: y = c₁ + c₂x + (c₃ + c₄x + c₅x2)e^x.
   (d) Multiple complex roots: If are complex double roots, the corresponding linearly independent solutions are:
   e^px cos(qx), e^px sin(qx),xe^px cos(qx), xe^px sin(qx), .
2. Convert the higher-order differential equation to a system of first-order equations.
   Example: .
   

(ii) Non - Homogeneous Linear ODEs

1.  y⁽ⁿ⁾ + p_n-1(x)y^(n-1) + ...+ p₁(x)y' + p₀(x)y = g(x)  the general solution is of the form: y = y_h + y_p where y_h is the homogeneous solution and y_p is a particular solution.
2.  Method of undermined coefficients
   Example: y'''+ 3y'' + 3y'+y = 30e
   Solution:  y = (c₁ + c₂x + c₃x²)e^-x + 5x³e^-x
3. Method of variation of parameters:
   y_{p = u1y1 +...+ unyn}
   where , k = 1,...., n.
   Example: .
   Solution: .

Points to Remember

1. F(x,y,y', ....y⁽ⁿ⁾) = 0, a nth order ODE if the nth derivative of the unknown function y(x) is the highest occurring derivative.
2. Linear ODE: y⁽ⁿ⁾ + p_{n - 1}(x)y^{(n - 1)} + ...+ p₁(x)y' + p₀(x)y = g(x)
3. Homogeneous linear ODE: y⁽ⁿ⁾ + p_{n - 1}(x)y^{(n - 1)} + ...+ p₁(x)y' + p₀(x)y = g(x) = 0
   Theorem: Fundamental Theorem for the Homogeneous Linear ODE: For a homogeneous linear ODE, sums and constant multiples of solutions on some open interval I are again solutions on I. (This does not hold for a nonhomogeneous or nonlinear ODE!).
   General solution: y= c₁y₁+...+ c_ny_n, where y₁,....,y_n is a basis (or fundamental system) of solutions on I; that is, these solutions are linearly independent on I.
4. Linear independence and dependence: n functions y₁,....,y_n are called linearly independent on some interval I where they are defined if the equation k_1y1,....,k_ny_{n = 0}  on I implies that all k₁,....,k_n are zero. These functions are called linearly dependent on I if this equation also holds on I for some k₁,....,k_n not all zero.
   Example: .  Sol.:  y = c₁e^-2x + c₂e^-x + c₃e^x + c₄e^2x
   Theorem: Let the homogeneous linear ODE have continuous coefficients p₀(x),... , p_n-1(x) on an open interval I. Then n solutions y₁,....,y_n on I are linearly dependent on I if and only if their Wronskian is zero for some x= x₀ in I. Furthermore, if W is zero for x= x₀, then W is identically zero on I. Hence if there is an in I at which W is not zero, then y₁,....,y_n are linearly independent on I, so that they form a basis of solutions of the homogeneous linear ODE on I.
   Wronskian:
5. Initial value problem: An ODE with n initial conditions y(x₀) =K₀, y'(x₀) = K₁, y_n-1(x₀) = K_n .

Example 1. Determine the order and degree of

Solution: The given differential equation when written as a polynomial in derivatives becomes

The highest order differential coefficient in this equation is and its power is 2.
The order is 2 and degree is 2.

Example 2. Solve dy/dx = (x + y + 1)2, if y(0) =0
Solution: Putting x + y + 1 = t, we get
Thus, the given equation becomes
Thus, the given equation becomes or
Integrating both side, we get

or tan^-1 t = x + c ⇒ tan-1(x + y + 1) = x + c
⇒ x + y + 1 = tan (x + c) When x = 0, y = 0 1 = tan (c)
⇒ c = p/4
Thus, the solution is given by x + y + 1 tan (x + p/4).

Example 3. Solve the differential equation (x² – y²) dx + 2xy dy = 0, given that y = 1 when x = 1
Solution: We have (x² – y²)dx + 2xy dy = 0
(x² – y²)dx = - 2xydy
...(i)

⇒ log ( v² + 1) = - log | x | +c
⇒ log(v² + 1) + log|x| = logc
⇒ (v² + 1)|x| = c
Now putting v = y/x
(y²/x² + 1)|x| = c
⇒ (x² + y²) = c|x|
Substituting x = 1 and y = 1, we get
c =2
Putting value of c = 2 in equation (2), we get
x² + y² = 2x or x² + y² = 2(-x).
Hence, x² + y² = 2x is the required solution.