Question

Solve the PDE p cos (x +y) + q sin (x + y) = z

Answer 1

Step-by-step explanation:

Lagrange's auxiliary equations are

dx/cos(x + y) = dy/sin(x + y) = dz/z ..... (1)

Taking 1, 1, 0 as multipliers, from (1), we get each ratio of (1)

= (dx + dy)/{cos(x + y) + sin(x + y)}

= d(x + y) / {cos(x + y) + sin(x + y)} ..... (2)

Taking 1, - 1, 0 as multipliers, from (1), we get each ratio of (1)

= (dx - dy)/{cos(x + y) - sin(x + y)}

= d(x - y) / {cos(x + y) - sin(x + y)} ..... (3)

From (1), (2), (3), we write

dz/z = d(x + y) / {cos(x + y) + sin(x + y)}

= d(x - y) / {cos(x + y) - sin(x + y)} ..... (4)

Taking the first two fractions of (4),

dz/z = d(x + y) / {cos(x + y) + sin(x + y)}

Let x + y = t. Then

dz/z = dt/(cost + sint)

= dt/{√2 * sin(π/4 + t)}

or, √2 * dz/z = cosec(π/4 + t) dt

On integration, we get

√2 * logz = log{tan(π/8 + 1/2)} + logc₁

or, z^(√2) = c₁ tan{π/8 + (x + y)/2}

or, z^(√2) * cot{π/8 + (x + y)/2} = c₁

Taking the last two fractions of (4) & let x + y = t.

Then,

d(x - y) = (cost - sint)/(cost + sint) dt

Integrating,

x - y = log(sint + cost) + logc

or, c * (sint + cost) = eˣ⁻ʸ

or, {sin(x + y) + cos(x + y)} * eʸ⁻ˣ = 1/c = c₂ (say)

Therefore the general solution can be written as

Φ [z^(√2) cot{π/8 + (x + y)/2}, {sin(x + y) + cos(x + y)} eʸ⁻ˣ] = 0

where Φ is an arbitrary function.

Note: Considered c, c₁, c₂ are constants.

Answer 2

Answer:

Step-by-step explanation:17.11.2018

Report

ANSWER

Answers

priyankajha2716 is waiting for your help.

Add your answer and earn points.

Swarup1998Samaritan

Step-by-step explanation:

Lagrange's auxiliary equations are

dx/cos(x + y) = dy/sin(x + y) = dz/z ..... (1)

Taking 1, 1, 0 as multipliers, from (1), we get each ratio of (1)

= (dx + dy)/{cos(x + y) + sin(x + y)}

= d(x + y) / {cos(x + y) + sin(x + y)} ..... (2)

Taking 1, - 1, 0 as multipliers, from (1), we get each ratio of (1)

= (dx - dy)/{cos(x + y) - sin(x + y)}

= d(x - y) / {cos(x + y) - sin(x + y)} ..... (3)

From (1), (2), (3), we write

dz/z = d(x + y) / {cos(x + y) + sin(x + y)}

= d(x - y) / {cos(x + y) - sin(x + y)} ..... (4)

Taking the first two fractions of (4),

dz/z = d(x + y) / {cos(x + y) + sin(x + y)}

Let x + y = t. Then

dz/z = dt/(cost + sint)

= dt/{√2 * sin(π/4 + t)}

or, √2 * dz/z = cosec(π/4 + t) dt

On integration, we get

√2 * logz = log{tan(π/8 + 1/2)} + logc₁

or, z^(√2) = c₁ tan{π/8 + (x + y)/2}

or, z^(√2) * cot{π/8 + (x + y)/2} = c₁

Taking the last two fractions of (4) & let x + y = t.

Then,

d(x - y) = (cost - sint)/(cost + sint) dt

Integrating,

x - y = log(sint + cost) + logc

or, c * (sint + cost) = eˣ⁻ʸ

or, {sin(x + y) + cos(x + y)} * eʸ⁻ˣ = 1/c = c₂ (say)

Therefore the general solution can be written as

If you think my answer is correct then click as brainiest