Test: First Order Differential Equations - 2 - Electronics and Communication Engineering (ECE) MCQ

It is in the standard form of first order linear equation.

Integrating factor

Solution of differential equation is:

If t = ±9, solution doesn’t exist.

Hence t ≠ ±9.

From the options, (-2, 2) doesn’t consists of ±9, hence, (-2, 2) is the correct.

Put x + y = v

Concept:

Variable separable: Equation of the form dy/dx = f(ax + by + c) can be reduced to variable separable form by putting ax + by + c = t
Calculation:
Given:

Substitute x + y – 1 = t

tan^-1 t = x + c

t = x + y – 1

tan^-1 (x + y – 1) = x + c

x + y – 1 = tan (x + c)

y = 1 – x + tan (x + c)

Concept:

• Differential equation: It is an equation that involves differential coefficients or differentials.
• Order of differential equation: It is the order of the highest derivative appearing in that equation.
• Degree of the differential equation: It is the degree (power) of the highest derivative occurring in that equation after the equation has been expressed in a form, free from radicals and fractional powers as far as the derivatives are concerned.

Explanation:

Given D.E is

Hence the highest order in the equation is 1 of dy/dx and the highest degree (power) of the equation is 2 of dy/dx

Hence the equation is First-order and Second-degree.

Concept:

If the differential equation is in the form of:

Integrating factor: 

IF = e^∫Pdx

Solution for equation:

y(IF) = ∫(IF)Qdx + C

Calculation:

where 0 ≤ x ≤ 1 and y (0) = 2.25

This differential equation is in the linear form,

Where P = 4, Q = 5

∴ IF = e^∫Pdx = e^∫4dx = e^4x

Solution of differential equation

Using y (0) = 2.25

c = 1

Concept:

Equation of the form y = x × g(p) + f(p) is called Lagrange's form.

When g(p) = p, then the equation, y = px + f(p) is called Clairaut's equation and the solution of such type of equation is given by: y = Cx + f(C).

Calculation:

Given:

The above equation represents Clairaut's form so the solution is y = Cx + f(C).

∴ (y - Cx)² = 4 + C²

∴ (y - Cx)² - C² = 4 is the required solution.

Concept:

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

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

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:

Given the differential equation is

 

Put tan y = t

By differentiating with respect to t,

⇒ sec² y dy = dt

Now, the given equation becomes

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

Integrating factor

Concept:

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

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

Calculation:

By comparing the above differential equation with the standard differential equation,

P = 1, Q = 2x

Integrating factor, IF = e^∫Pdx = e^∫1dx = e^x

Now, the solution is

y(e^x) = ∫e^x(2x)dx

ye^x = 2(xe^x − e^x) + C

⇒ y = 2(x − 1) + Ce^−x

y(0) = 1

⇒ 1 = 2 (0 – 1) + C

⇒ C = 3

Now, the solution becomes

y = 2x – 2 + 3e^-x

At x = ln 2,

y = 2 ln 2 – 2 + 3 (0.5) = 0.8862

Concept:

Equation of this type is known as Linear First Order Equation, whose solution is given by: y(I.F.) = ∫(Q × I.F.)dx + c whereI.F. = e^∫Pdx

Calculation:

Given:

By comparing equation (1) with

Solving through partial fraction method

(x + 1)(x - 1)A + x(x-1)B + x(x + 1)C = 4x² – 2

Ax² – A + Bx² – Bx + Cx² + Cx = 4x² – 2

(A + B + C)x² + (C - B)x – A =  4x² – 2

Comparing co-efficient of x², x, and constants we get

A + B + C = 4 ……. (ii)

C – B = 0 

⇒ B = C

A = 2

∵ A + B + C = 4

⇒ 2 + B + C = 4

⇒ B + C = 2

∴ B = C = 1

⇒ 2ln x + ln (x + 1) + ln (x - 1)

⇒ ln x² + ln (x² - 1)

⇒ ln x²(x² - 1)

∵I.F. = e^∫Pdx

⇒ I.F = x² (x² - 1)

Concept:

For solving first order, first-degree differential equations always first inspect with variable separation method.

Calculation:

Given the differential equation is,

 separating variables

dy/y = −7x²dx, integrating both sides;

Where A is a constant.

Use condition y(0) = 3/7 in (1)