ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.5

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.5

Solve the following (1 to 5) differential equations:

Question 1.
(i) (x + y)² \(\frac{d y}{d x}\) = 1
(ii) \(\frac{d y}{d x}\) = (4x + y + 1)²
Solution:
(i) Given (x + y)² \(\frac{d y}{d x}\) = 1
putting x + y = t
⇒ 1 + \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\)
⇒ \(\frac{d y}{d x}\) = \(\frac{d y}{d x}\) – 1
∴ from eqn. (1) ; we have
t² [\(\frac{d y}{d x}\) – 1] = 1
⇒ \(\frac{d t}{d x}\) = \(\frac{1}{t^2}\) + 1
⇒ \(\frac{d t}{d x}=\frac{1+t^2}{t^2}\)
⇒ \(\frac{t^2 d t}{t^2+1}\) = dx
on integrating ; we have
∫ \(\left[\frac{t^2+1-1}{t^2+1}\right]\) dt = ∫ dx + c
⇒ ∫ [1 – \(\frac{1}{t^2+1}\)] dt = x + c
⇒ t – tan^-1 t = x c
⇒ x + y – tan^-1 (x + y) = x + c
⇒ y – tan^-1 (x + y) = c be the required solution.

(ii) Given eqn. is
\(\frac{d y}{d x}\) = (4x + y + 1)²
put 4x + y + 1 = t ;
Diff. w.r.t. x, we have
4 + \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\)
⇒ \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\) – 4
∴ From (1) ; we have
\(\frac{d t}{d x}\) – 4 = t²
⇒ \(\frac{d t}{d x}\) = t² + 4
⇒ \(\frac{d t}{t^2+4}\) = dx
[seperation of variables]
On integrating ; we have
\(\frac{1}{2}\) tan^-1 (\(\frac{t}{2}\)) = x + C
⇒ \(\frac{1}{2} \tan \left[\frac{4 x+y+1}{2}\right]\) = x + C is the required solution.

Question 2.
(i) (x – y)² \(\frac{d y}{d x}\) = a²
(ii) \(\frac{d y}{d x}\) = tan² (x + y).
Solution:
(i) Given Diff. eqn. is
(x – y)² \(\frac{d y}{d x}\) = a² ……………….(1)
put x – y =t ;
Diff. w.r.t. x, we get

(ii) Given, \(\frac{d y}{d x}\) = tan² (x + y) …………….(1)
put x + y = t
⇒ 1 + \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\)
⇒ \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\) – 1
∴ from (1) ;
\(\frac{d t}{d x}\) – 1 = tan² t
⇒ \(\frac{d t}{d x}\) = 1 + tan² t = sec² t
On variable seperation, we have
∫ \(\frac{d t}{\sec ^2 t}\) = ∫ dx + C
⇒ ∫ cos² t = ∫ dx + C
⇒ ∫ \(\left[\frac{1+\cos 2 t}{2}\right]\) dt = x + C
⇒ \(\frac{1}{2}\left[t+\frac{\sin 2 t}{2}\right]\) = x + C
⇒ x + y + \(\frac{1}{2}\) sin 2 (x + y) = 2x + A
⇒ y – x + \(\frac{1}{2}\) sin 2 (x + y) = A
⇒ 2 (y – x) + sin 2 (x + y) = 2A = A’
which is the required solution.

Question 3.
(i) \(\frac{d y}{d x}\) = (3x + y + 4)²
(ii) cos (x + y) dy = dx
Solution:
(i) Given \(\frac{d y}{d x}\) = (3x + y + 4)² ……………….(1)
put 3x + y + 4 = t
⇒ 3 + \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\)
[Diff. both sides w.r.t x]
⇒ \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\) – 3
∴ from (1) ;
\(\frac{d t}{d x}\) – 3 = t²
⇒ \(\frac{d t}{d x}\) = t² + 3
⇒ \(\frac{d t}{t^2+3}\) = dx
[after variable separation]
On integrating ; we have

(ii) Given eqn. is cos (x + y) dy = dx
⇒ cos (x + y) = \(\frac{d x}{d y}\)
⇒ \(\frac{d y}{d x}\) = sec (x + y) ……………….(1)
put x + y = t ;
Diff. w.r.t. x, we get
1 + \(\frac{d y}{d x}\) = cos (x + y)
(ii) cos = \(\frac{d t}{d x}\)
⇒ \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\) – 1,
putting in eqn. (1)
\(\frac{d t}{d x}\) – 1 = sec t
⇒ \(\frac{d t}{d x}\) = sec t + 1
⇒ \(\frac{d t}{1+\sec t}\) = dx ;
On integrating

Question 4.
(i) \(\frac{d y}{d x}\) = cos (x + y)
(ii) cos² (x – 2y) = 1 – 2 \(\frac{d y}{d x}\).
Solution:
(i) Given \(\frac{d y}{d x}\) = cos (x + y) ……………….(1)
put x + y = t
⇒ 1 + \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\)

(ii) Given cos² (x – 2y) = 1 – 2 \(\frac{d y}{d x}\)
put x – 2y = t
⇒ 1 – 2 \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\)
∴ From eqn. (1) ; we have
cos² t = \(\frac{d t}{d x}\)
⇒ dx = \(\frac{d t}{\cos ^2 t}\) ;
on integrating ; we have
∫ dx = ∫ sec² t dt
⇒ x = tan t + c
⇒ x = tan (x – 2y) + c be the required solution.

Question 5.
(i) (x + y + 1) \(\frac{d y}{d x}\) = 1
(ii) \(\frac{d y}{d x}=\frac{x+y-1}{x+y+1}\)
Solution:
(i) Given (x + y + 1) \(\frac{d y}{d x}\) = 1
⇒ \(\frac{d y}{d x}=\frac{1}{x+y+1}\) …………………(1)
put x + y + 1 = t
⇒ 1 + \(\frac{d y}{d x}=\frac{d t}{d x}\)
⇒ \(\frac{d y}{d x}=\frac{d t}{d x}\) – 1
Thus from eqn. (1) ; we have
\(\frac{d t}{d x}\) – 1 = \(\frac{1}{t}\)
\(\frac{d t}{d x}\) = 1 + \(\frac{1}{t}\)
= \(\frac{t + 1}{t}\)) dt = 2 dx
⇒ \(\frac{t d t}{t+1}\) = dx ;
on integrating
∫ \(\left[1-\frac{1}{t+1}\right]\) dt = ∫ dx
⇒ t – log (t + 1) = x + c
⇒ x + y + 1 – log (x + y + 2) = x + c
⇒ log (x + y + 2) = y + 1 -c
⇒ log (x + y + 2) = y + c’
⇒ x + y +2 = e^{y + c’}
⇒ x = Ae^y – y – 2 be the required solution.

(ii) Given \(\frac{d y}{d x}=\frac{x+y-1}{x+y+1}\) ………………….(1)
put x + y = t;
Diff., both sides w.r.t. x
1 + \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\)
∴ from eqn. (1) ;
\(\frac{d t}{d x}\) – 1 = \(\frac{t-1}{t+1}\)
⇒ \(\frac{d t}{d x}=\frac{t-1}{t+1}+1=\frac{2 t}{t+1}\)
⇒ (\(\frac{t + 1}{t}\)) dt = 2 dx
on integrating ; we get
⇒ ∫ (1 + \(\frac{1}{t}\)) dt = 2 ∫ dx + C
⇒ t + log |t| = 2x + C
⇒ x + y + log (x + y) = 2x + C
⇒ y – x + log |x + y| = C
which is the required solution.

Question 6.
Find a particular solution of the differential equation (x + y + 1)² dy = dx, given that y = 0 when x = – 1.
Solution:
Given diff. eqn. can be written as
(x + y + 1)² = 1 ……………..(1)
putting x + y + 1 = t
⇒ 1 + \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\)
[Diff. both sides w.r.t. x]
⇒ \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\) – 1
∴ from (1) ;
t² (\(\frac{d t}{d x}\) – 1) = 1
⇒ \(\frac{d t}{d x}=\frac{1}{t^2}\) + 1
⇒ \(\frac{d t}{d x}=\frac{t^2+1}{t^2}\)
⇒ \(\frac{t^2 d t}{t^2+1}\) = dx
On integrating; we have
∫ \(\left[\frac{t^2+1-1}{t^2+1}\right]\) dt = ∫ dx + C
⇒ ∫ [1 – \(\frac{1}{t^2+1}\)] dt = x + C
⇒ t – tan^-1 t = x + C
⇒ x + y + 1 – tan^-1 (x + y + 1) = x + C
⇒ y + 1 – tan^-1 (x + y + 1) = C …………….(2)
which gives the general solution of given differential equation.
For particular solution of eqn. (1) ; we have
given that y = 0
when x = – 1
∴from (2) ; we get
0 + 1 – tan^-1 (- 1 + 0 + 1) C
⇒ C = 1
Thus eqn. (2) becomes ;
y + 1 – tan^-1 (x + y + 1) = 1
⇒ tan^-1 (x + y + 1) = y
⇒ x + y + 1 = tan y be the required solution.