Test: Method of Parameter Variations - Civil Engineering (CE) MCQ

(x² – y²) dx – xydy = 0

This homogeneous in x & y

Separating the variables

Integrating on both side

⇒ 4logx + log⁡(1 − 2v²) = −4c

⇒ x² (x² – 2y²) = constant

Concept:

Calculation:

Given:

(D² – 6D + 9)y = e^3x/x²

Auxiliary equation is

(D² – 6D + 9) = 0

⇒ (D – 3)² = 0

C.F. = (C₁ + C₂x) e^3x

C.F = C₁(e^3x) + C₂(xe^3x)

Auxiliary equation, D² – 1 = 0

⇒ D = ± 1

C.F. = C₁e^x + C₂e^−x

Here, y₁ = e^x

y₂ = e^-x

y₁′ = e^x
y₂′ = −e^−x

= -1 - 1 = -2

If the given differential equation is in the form, y'' + py' + qy = x

Where, p, q, x are functions of x.

Then Wronskian is,

Here, y₁ and y₂ are the solutions of y'' + py' + qy = 0

Given differential equation is,

Auxiliary equation is,

D² – 3D + 2 = 0

⇒ D = 1, 2

C.F. = c₁ e^2x + c₂ e^x

y₁ = e^2x, y₂ = e^x

Given differential eqaution is,

Let, y = v × x

dy = vdx + xdv

By substituting values of Y and dY in equation 1, we get

Integrating both sides

2 log x = log |sec v| - log v + log c

Give equation is (D² – 6D + 9)y = e^3x/x²

Auxiliary equation is, (D² – 6D + 9) = 0

⇒ (D – 3)² = 0

C.F. = (C₁ + C₂x) e^3x

Given differential equation is

auxiliary equation is (D − 2)² = 0 ⇒ D = 2, 2

This problem can be solved by using the method of variation of parameters. Then

Auxiliary equation:

m² + 4 = 0

m = ±2i

By wrongskion,

PI = A u(x) + B v(x)