(a) Show that the general solution of the differential equation dy/dx + Py = Q can be written in the form y = Q/P - e⁻∫ P dxC + ∫…

(a) Let I = ∫ P dx. The equation is dy/dx + P y = Q. Using the integrating-factor method, multiply by exp(I):

exp(I) dy/dx + P y exp(I) = Q exp(I).

Since d(exp(I))/dx = P exp(I),

d/dx (y exp(I)) = Q exp(I).

Integrating,

y exp(I) = C + ∫ Q exp(I) dx.

Hence

y = exp(-I) [C + ∫ Q exp(I) dx].

Now Q exp(I) dx = (Q/P) d(exp(I)), because d(exp(I)) = P exp(I) dx. Therefore, by integration by parts,

∫ Q exp(I) dx = ∫ (Q/P) d(exp(I)) = (Q/P) exp(I) - ∫ exp(I) d(Q/P).

Substituting,

y = exp(-I) [C + (Q/P) exp(I) - ∫ exp(I) d(Q/P)].

Thus

y = Q/P + C exp(-I) - exp(-I) ∫ exp(I) d(Q/P).

Since C is arbitrary, replace C by -C. Hence

y = Q/P - exp(-I) [C + ∫ exp(I) d(Q/P)],

that is,

y = Q/P - e^(-∫ P dx) [C + ∫ e^(∫ P dx) d(Q/P)].

This is the required form. The result requires P ≠ 0 so that Q/P is defined.

Final answer: y = Q/P - e^(-∫ P dx) [C + ∫ e^(∫ P dx) d(Q/P)].

(b) The system of parabolas is

x² = 4a(y + a).

Differentiate with respect to x:

2x = 4a dy/dx.

Let p = dy/dx. Then

a = x/(2p).

Substitute this value of a into the original equation:

x² = 4 (x/(2p)) (y + x/(2p)).

Therefore,

x² = 2xy/p + x²/p².

Multiplying by p²,

x² p² = 2xy p + x².

Dividing by x²,

p² = (2y/x) p + 1.

Thus the differential equation of the given system is

p² - (2y/x) p - 1 = 0.

For an orthogonal trajectory, the slope m satisfies m = -1/p, i.e. p = -1/m. Substitute this into the differential equation:

1/m² = (2y/x)(-1/m) + 1.

Multiplying by m²,

1 = -(2y/x) m + m².

Hence

m² - (2y/x) m - 1 = 0.

This is exactly the same differential equation as the original system. Therefore the orthogonal trajectories satisfy the same differential equation and hence belong to the same family of parabolas, possibly with a different parameter.

Final answer: The orthogonal trajectories belong to the same system x² = 4a(y + a).

(c) Let the weight of the body be w. On the inclined plane of inclination θ, the normal reaction is

N = w cosθ.

Hence the limiting friction is

F_f = μ N = μ w cosθ.

It is given that μ > tanθ, so

μ cosθ > sinθ.

While dragging the body up the plane, friction acts down the plane. Therefore the applied force up the plane is

F₁ = w sinθ + μ w cosθ.

Work done in dragging it a distance b up the plane is

W₁ = F₁ b = w b (sinθ + μ cosθ).

While dragging it back down the plane, friction acts up the plane. Since μ cosθ > sinθ, the applied force must act down the plane to overcome the excess friction. Thus

F₂ = μ w cosθ - w sinθ.

Work done in dragging it back a distance b down the plane is

W₂ = F₂ b = w b (μ cosθ - sinθ).

Total work done by the applied force is

W = W₁ + W₂ = w b (sinθ + μ cosθ) + w b (μ cosθ - sinθ).

Therefore,

W = 2 μ w b cosθ.

The gravitational work cancels over the upward and downward journeys, while friction opposes motion in both directions.

Final answer: W = 2 μ w b cosθ. Units: joule if w is in newton and b in metre.

(d) Let the projectile speed be

u = √(2gh).

Measure θ positive below OX (the direction of OY), and write m = tanθ. Since OY is vertically downward, the equations of motion are

x = u cosθ t,

y = u sinθ t + (1/2) g t².

Eliminating t = x/(u cosθ),

y = x tanθ + (g x²)/(2u²) sec²θ.

Since sec²θ = 1 + tan²θ,

y = x m + (g x²)/(2u²) (1 + m²).

But u² = 2gh, so

g/(2u²) = g/(4gh) = 1/(4h).

Hence

y = x m + (x²/(4h)) (1 + m²).

Multiplying by 4h,

4h y = 4h x m + x² + x² m².

Thus

x² m² + 4h x m + x² - 4h y = 0.

This quadratic in m gives the two possible directions of projection. If their directions are at right angles, then

m₁ m₂ = -1.

The product of the roots of the quadratic is

m₁ m₂ = (x² - 4h y)/x².

Therefore,

(x² - 4h y)/x² = -1.

So

x² - 4h y = -x²,

which gives

2x² = 4h y,

and hence

x² = 2h y.

Now let φ = ∠POX. Since P = (x, y),

tanφ = y/x.

Using x² = 2h y, we get

tanφ = y/x = x/(2h).

Let t = tan(φ/2). Then

tanφ = 2t/(1 - t²).

Thus

2t/(1 - t²) = x/(2h).

Cross-multiplying,

4h t = x(1 - t²) = x - x t²,

so

x t² + 4h t - x = 0.

But under x² = 2h y, the projectile quadratic becomes

x² m² + 4h x m + x² - 4h y = 0.

Substitute 4h y = 2x²:

x² m² + 4h x m - x² = 0.

Dividing by x,

x m² + 4h m - x = 0.

This is the same quadratic as the one satisfied by t = tan(φ/2). Therefore one possible value of m is tan(φ/2). Hence one possible direction of projection makes angle φ/2 with OX, i.e. it bisects ∠POX.

Final answer: x² = 2h y, and one possible direction of projection bisects ∠POX.

(e) For the vector field

A = (6xy + z³) i + (3x² - z) j + (3xz² - y) k,

compute ∇ × A. Its components are:

(∇ × A)_x = ∂A_z/∂y - ∂A_y/∂z.

Here

∂A_z/∂y = ∂/∂y (3xz² - y) = -1,

and

∂A_y/∂z = ∂/∂z (3x² - z) = -1.

Thus

(∇ × A)_x = -1 - (-1) = 0.

Next,

(∇ × A)_y = ∂A_x/∂z - ∂A_z/∂x.

Now

∂A_x/∂z = ∂/∂z (6xy + z³) = 3z²,

and

∂A_z/∂x = ∂/∂x (3xz² - y) = 3z².

Hence

(∇ × A)_y = 3z² - 3z² = 0.

Finally,

(∇ × A)_z = ∂A_y/∂x - ∂A_x/∂y.

Now

∂A_y/∂x = ∂/∂x (3x² - z) = 6x,

and

∂A_x/∂y = ∂/∂y (6xy + z³) = 6x.

Therefore,

(∇ × A)_z = 6x - 6x = 0.

Thus

∇ × A = 0.

So A is irrotational.

Now find φ such that

A = ∇φ.

Thus

∂φ/∂x = 6xy + z³.

Integrating with respect to x,

φ = 3x² y + x z³ + f(y, z).

Differentiate with respect to y:

∂φ/∂y = 3x² + ∂f/∂y.

But this must equal A_y = 3x² - z. Hence

∂f/∂y = -z.

Integrating with respect to y,

f = -y z + g(z).

So

φ = 3x² y + x z³ - y z + g(z).

Differentiate with respect to z:

∂φ/∂z = 3x z² - y + g'(z).

This must equal A_z = 3x z² - y. Therefore,

g'(z) = 0,

so g(z) = C, a constant.

Hence

φ = 3x² y + x z³ - y z + C.

Final answer: A is irrotational, and φ = 3x² y + x z³ - y z + C.