Chapter 6 Differential Equations Ex 6.3 Solutions

Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 6 Differential Equations Ex 6.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

(ii) y = (sin^-1x)²+ c; (1 – x²)
Solution:
y = (sin^-1x)²+ c …….(1)
Differentiating w.r.t. x, we get

Differentiating again w.r.t. x, we get

(iii) y = e^-x+ Ax + B;
Solution:
y = e^-x+ Ax + B
Differentiating w.r.t. x, we get

∴
Hence, y = e^-x+ Ax + B is a solution of the D.E.

(iv) y = x^m;
Solution:
y = x^m
Differentiating twice w.r.t. x, we get

This shows that y = x^mis a solution of the D.E.

(v) y = a + ;
Solution:
y = a +
Differentiating w.r.t. x, we get

Differentiating again w.r.t. x, we get

Hence, y = a + is a solution of the D.E.

(vi) y = e^ax; x = y log y
Solution:
y = e^ax
log y = log e^ax= ax log e
log y = ax …….(1) ……..[∵ log e = 1]
Differentiating w.r.t. x, we get
= a × 1
∴ = ay
∴ x = (ax)y
∴ x = y log y ………[By (1)]
Hence, y = e^axis a solution of the D.E.
x = y log y.

(ii) log() = 2x + 3y
Solution:

(iii) y – x = 0
Solution:
y – x = 0
∴ x = y
∴
Integrating both sides, we get

∴ log |x| = log |y| + log c
∴ log |x| = log |cy|
∴ x = cy
This is the general solution.

(iv) sec²x. tan y dx + sec²y. tan x dy = 0
Solution:
sec²x. tan y dx + sec²y. tan x dy = 0
∴
Integrating both sides, we get

Each of these integrals is of the type
= log |f(x)| + c
∴ the general solution is
∴ log|tan x| + log|tan y | = log c, where c₁= log c
∴ log |tan x. tan y| = log c
∴ tan x. tan y = c
This is the general solution.

(v) cos x. cos y dy – sin x. sin y dx = 0
Solution:
cos x. cos y dy – sin x. sin y dx = 0

Integrating both sides, we get
∫cot y dy – ∫tan x dx = c₁
∴ log|sin y| – [-log|cos x|] = log c, where c₁= log c
∴ log |sin y| + log|cos x| = log c
∴ log|sin y. cos x| = log c
∴ sin y. cos x = c
This is the general solution.

(vi) = -k, where k is a constant.
Solution:
= -k
∴ dy = -k dx
Integrating both sides, we get
∫dy = -k∫dx
∴ y = -kx + c
This is the general solution.

(vii)
Solution:

∴ y cos²y dy + x cos²x dx = 0
∴
∴ x(1 + cos 2x) dx + y(1 + cos 2y) dy = 0
∴ x dx + x cos 2x dx + y dy+ y cos 2y dy = 0
Integrating both sides, we get
∫x dx + ∫y dy + ∫x cos 2x dx + ∫y cos 2y dy = c₁……..(1)
Using integration by parts

Multiplying throughout by 4, this becomes
2x²+ 2y²+ 2x sin 2x + cos 2x + 2y sin 2y + cos 2y = 4c₁
∴ 2(x²+ y²) + 2(x sin 2x + y sin 2y) + cos 2y + cos 2x + c = 0, where c = -4c₁
This is the general solution.

(viii)
Solution:

(ix) 2e^x+2ydx – 3 dy = 0
Solution:

(x) = e^x+y+ x²e^y
Solution:

∴ 3e^x+ 3e^-y+ x³= -3c₁
∴ 3e^x+ 3e^-y+ x³= c, where c = -3c₁
This is the general solution.

(ii) (x – y²x) dx – (y + x²y) dy = 0, when x = 2, y = 0
Solution:
(x – y²x) dx – (y + x²y) dy = 0
∴ x(1 – y²) dx – y(1 + x²) dy = 0

When x = 2, y = 0, we have
(1 + 4)(1 – 0) = c
∴ c = 5
∴ the particular solution is (1 + x²)(1 – y²) = 5.

(iii) y(1 + log x) – x log x = 0, y = e², when x = e
Solution:
y(1 + log x) – x log x = 0

(iv) (e^y+ 1) cos x + e^ysin x = 0, when x = , y = 0
Solution:
(e^y+ 1) cos x + e^ysin x = 0

= log|f(x)| + c
∴ from (1), the general solution is
log|sin x| + log|e^y+ 1| = log c, where c₁= log c
∴ log|sin x. (e^y+ 1)| = log c
∴ sin x. (e^y+ 1) = c
When x = , y = 0, we get

∴ c = (1 + 1) = √2
∴ the particular solution is sin x. (e^y+ 1) = √2

(v) (x + 1) – 1 = 2e^-y, y = 0, when x = 1
Solution:

This is the general solution.
Now, y = 0, when x = 1
∴ 2 + e⁰= c(1 + 1)
∴ 3 = 2c
∴ c =
∴ the particular solution is 2 + e^y= (x + 1)
∴ 2(2 + e^y) = 3(x + 1).

(vi) cos() = a, a ∈ R, y (0) = 2
Solution:
cos() = a
∴ = cos^-1a
∴ dy = (cos^-1a) dx
Integrating both sides, we get
∫dy = (cos^-1a) ∫dx
∴ y = (cos^-1a) x + c
∴ y = x cos^-1a + c
This is the general solution.
Now, y(0) = 2, i.e. y = 2,
when x = 0, 2 = 0 + c
∴ c = 2
∴ the particular solution is
∴ y = x cos^-1a + 2
∴ y – 2 = x cos^-1a
∴ = cos^-1a
∴ cos() = a

(ii) (x – y)² = a²
Solution:

(iii) x + y = sec(x²+ y²)
Solution:

Integrating both sides, we get
∫cos u du = 2 ∫dx
∴ sin u = 2x + c
∴ sin(x²+ y²) = 2x + c
This is the general solution.

(iv) cos²(x – 2y) = 1 – 2
Solution:

Integrating both sides, we get
∫dx = ∫sec²u du
∴ x = tan u + c
∴ x = tan(x – 2y) + c
This is the general solution.

(v) (2x – 2y + 3) dx – (x – y + 1) dy = 0, when x = 0, y = 1
Solution:
(2x – 2y + 3) dx – (x – y + 1) dy = 0
∴ (x – y + 1) dy = (2x – 2y + 3) dx
∴ ………(1)
Put x – y = u, Then

∴ u – log|u + 2| = -x + c
∴ x – y – log|x – y + 2| = -x + c
∴ (2x – y) – log|x – y + 2| = c
This is the general solution.
Now, y = 1, when x = 0.
∴ (0 – 1) – log|0 – 1 + 2| = c
∴ -1 – o = c
∴ c = -1
∴ the particular solution is
(2x – y) – log|x – y + 2| = -1
∴ (2x – y) – log|x – y + 2| + 1 = 0