Maths Quiz on Differential Equations for IIT JEE & Engineering Exam

b) 3

Solution:

x dy = y (dx + y dy)

[x dy − ydx] / [y²] = dy

−d (x / y) = dy

Integrating both sides, we get x / y + y = c [Because y (1) = 1 ⇒ c = 2; Hence xy + y = 2 ]

For x = −3,

y² − 2y − 3 = 0

⇒ y = −1 or 3

⇒y = 3 (Because y > 0)

b) ± √3e

Solution:

(x² + y²) dy = xy dx

x (x dy − y dx) = −y² dy

[x * (y dx − x dy)] / y² = dy

[x / y] d (x / y) = dy / y

Integrating, x² / 2y² = loge y + c

Given y (1) = 1

c = 1 / 2

x² / 2y² = loge y + 1 / 2

Now y (x₀) = e

x₀² / 2e² − loge e − 1 / 2 = 0

x₀² = 3e²

x₀ = ± √3e

d) x² + y² = px³

It is homogeneous equation dy / dx = [x² + 3y²] / 2xy

Put y = vx and dy / dx = v + x * [dv / dx]

So, we get x * [dv / dx] = [1 + v²] / 2v

2v dv / 1 + v² = dx / x

On integrating, we get

x² + y² = px³. (where p is a constant)

a) e^tan−1y

Solution:

(1 + y²) dx − (tan⁻¹ y − x) dy = 0

dy / dx = (1 + y²) / (tan⁻¹ y − x)

dx / dy = [tan⁻¹ y / 1 + y²] − [x / 1 + y²] [dx / dy] + [x / 1 + y²] = [tan⁻¹ y] / [1 + y²]

This is equation of the form dx / dy + Px = Q

So, I.F. = e^∫Pdy

= e^{∫1 / 1 + y2.dy}

= e^tan−1y

Solution:

(2x −y + 1) dx + (2y − x + 1) dy = 0

dy / dx = 2x − y + 1x − 2y −1, put x = X + h, y = Y + k

dY / dX = [2X − Y + 2h − k + 1] / [X − 2Y + h −2k − 1]

2h − k + 1 = 0

h − 2k − 1 = 0

On solving h = −1, k = −1;

dY/ dX = [2X − Y] / [X − 2Y]

Put Y = vX;

dY / dX = v + [X dv / dX]

v + [X dv / dX] = [2X − vX] / [X − 2vX] = [2 − v] / [1 − 2v]

X dv / dX = [2 − 2v + 2v²] / [1 − 2v] = 2 (v² −v + 1) / [1 − 2v]

dX / X = (1 − 2v) / 2 (v² − v + 1) dv

Put v² − v + 1 = t

(2v − 1) dv = dt

dX / X = −dt / 2t

log X = logt^{− ½} + log c

X = t^−1/2 c

X = (v² − v + 1)^−1/2 * c

X² (v² − v + 1) = constant

(x + 1)² (([y + 1)² / (x + 1)²] − [(y+1) / (x + 1)] + 1) = constant

(y + 1)² − (y + 1) (x + 1) + (x + 1)² = c

y² + x² − xy + x + y = c

Solution:

Given dy / dx = x −y / x + y.

Put y = vx

dy / dx = v + x * [dv / dx]

v + x * [dv / dx] = [x − vx] / [x + vx]

v + x [dv / dx] = [1 − v] / [1 + v] [1 + v] / [2 − (1 + v)²] dv = dx / x

Integrating both sides, ∫[1 + v] / [2 − (1 + v)²] dv = ∫dx / x

Put (1 + v)² = t

⇒ 2 (1 + v) dv = dt

[1 / 2] ∫dt / [2 − t] =∫dx / x

[− 1 / 2] log (2 − t) = log xc

[−1 / 2] log [2 − (1 + v)²] = log xc

[−1 / 2] log [−v² − 2v + 1] = log xc

log 1 / √[1 − 2v −v²] = log xc

x²c² (1 − 2v −v²) = 1

y² + 2xy − x² = c₁.

c) (x − 1)³

Solution:

Given f′′(x) = 6 (x − 1)

f′(x) = 3 (x − 1)² + c₁ ..(i)

But at point (2, 1) the line y = 3x − 5 is tangent to the graph y = f(x).

Hence, dy / dx∣_{x = 2} = 3 or

f′ (2) = 3.

Then from (i) f′ (2) = 3 (2 − 1)² + c₁

3 = 3 + c₁

c₁ = 0 i.e.,

f′ (x) = 3 (x − 1)²

Given f (2) = 1

f (x) = (x − 1)³ + c₂

f (2) = 1 + c₂

1 = 1 + c₂

c₂ = 0

Hence, f (x) = (x − 1)³.

Solution:

(x + y) dx + x dy = 0

x dy = − (x + y) dx

dy / dx = −[x + y] / x

It is homogeneous equation, hence, put y = vx and dy / dx = v + x [dv / dx], we get

v + x dv / dx = −[x + vx] / x= −[1 + v] / 1

x dv / dx = −1 −2v

∫dv / [1 + 2v] = −∫dx / x

[1 / 2] log (1 + 2v) = −log x + log c

log (1 + 2 [y / x]) = 2 log [c / x] [x + 2y] / x = (c / x)²

x² + 2xy = c.

d) 1

We have [dx / x] = [y dy] / [1 + y²]

Integrating, we get log |x| = ([1 / 2] * log [1 + y²]) + log c

|x| = c √(1 + y²)−−−−−−−

But it passes through (1, 0), so we get c = 1

Therefore, the solution is x² = y² + 1 or x² − y² = 1.

 a) x²

Slope dy / dx = 2y / x

2 ∫dx / x =∫dy / y

2 log x = log y + log c

x² = yc

Since it passes through (1, 1), therefore, c = 1.

Hence, x² − y = 0

y = x².