Test: First Order Differential Equations - 1 - Electronics and Communication Engineering (ECE) MCQ

Concept:

• 
  where ‘x’ is an algebraic function
• 
• 
• The second derivative is represented as
• 

Calculation:

Given:

y = sin x

Now second derivative of sin(x) with respect to x is

Concept:

Calculation:

Using: cos a – cos b = - 2 sin [(a + b)/2] × sin [(a - b)/2]

Concept:

Linear Differential Equation: Linear differential equations are those in which the dependent variable, its derivatives occur only in the first degree and they are not multiplied together. It is of the form:

Where, k₁, k₂, …k_n are the constants.

The solution of the equation is given as:

y = C.F 

where C.F is the complementary function.

The above linear differential equation in the symbolic form is represented as

(Dⁿ + k₁ D^n-1 + k₂ D^n-2 +…+ k_n) y = 0

For different roots of the auxiliary equation, the solution (complementary function) of the differential equation is as shown below.

Calculation:

(D − 0.2)i(t) = 0
D = 0.2
i(t) = k.e^0.2t, −10 < t < 10

At t = 4; 10 = K.e^0.8

K = 4.493
∴ i(−5) = 4.493 × e⁻¹
i(−5) = 1.65

At, n = -1

Integrating

ln y = -ln x + ln c

Ln (xy) = ln c

xy = c

at, n = +1

Ydy = -xdx

Integrating

x² + y² = c

Y (1) = -?

∵ D²y + Dy - 6y = 0

Characteristic equation will be -

m² + m - 6 = 0

m² + 3m - 2m - 6 = 0

m = -3, 2

Since roots are real & different thus:

y(x) = C₁^e−3x + C₂e^2x

Given y (0) = 0

0 = C₁ + C₂      ----(1)

1 = -3C₁ + 2C₂      ----(2)

By equation (1) & (2)

5C₁ = -1


= 1.467

Y (1) = 1.47

Concept:
Linear Partial Differential Equation of First Order: A linear partial differential equation of the first order, commonly known as Lagrange's Linear equation, is of the form Pp + Qq = R where P, Q and R are functions of x, y, z. This equation is called a quasi-linear equation. When P, Q and R are independent of z it is known as linear equation.
Thus, to solve the equation of the form Pp + Qq = R

1. form the subsidiary equations as dx/P = dy/Q = dz/R
2. solve these simultaneous equations by any the method giving u = a and v = b as its solutions.
3. write the complete solution as φ (u, v) = 0 or u = f (v).

Calculations:

Given partial differential equation, x(y − z)p + y(z − x)q = z(x − y)
Hence the subsidiary equation is,
So, using the property,, it can be written as 

 

∴ dx + dy + dz = 0  and by integrating we get, x + y + z = a

For the second function, using multipliers as 1/x, 1/y, 1/z and using the property,we get,

 and by integrating we get, lnx + lny + lnz = lnb ⇒ xyz = b

Hence, the solution is φ(u, v) = 0 ⇒ φ(x + y + z, xyz) = 0

This is single order differential equation of:

then  P = 1/x and Q = x³

then solution of equation

⇒ y(IF) = ∫Q(IF)dx + C

Applying boundary condition y(1) = 6/5

sec²ydy = cosec²x.dx
tan⁡ y = −cot ⁡x + c
tan⁡ y + cot ⁡x = c

y² = t

The above is linear differential equation of the form

Integrating factor

Solution is given by:

Substitute t = y²

At, x = 1, y = 0

0 = -e^-1 + c

c = e^-1 = y­_e

substituting in the solution

Taking log both sides

Concept:

The standard form of a first-order linear differential equation is,

Form 1: 

Where P and Q are the functions of x.

Integrating factor, IF = e^∫Pdx

Now, the solution for the above differential equation is,

y(IF) = ∫IF.Qdx + C

Form 2: 

Where P and Q are the functions of y.

Integrating factor, IF = e^∫Pdy

Now, the solution for the above differential equation is,

x(IF) = ∫IF.Qdy + C

Calculation:

The given differential equation is,

(1 + y²) dx = (tan^-1 y - x) dy

Now, it is in the standard form of a first-order linear differential equation.

Integrating factor:

Now, the solution of the given differential equation is

Now, the solution becomes