ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.2

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.2

Very short answer type questions (1 to 7) :

Question 1.
What is the number of arbitrary constants in the general solution of a differential equation of order 4 ? (NCERT)
Solution:
Since the general solution of the differential equation of order n contains n independent arbitrary constants.
Thus, the general solution of given diff. eqn. of order 4 contains n arbitrary constants.

Question 2.
What is the number of arbitrary constants in a particular solution of a differential equation of order 3 ? (NCERT)
Solution:
Since the particular solution of the diff. eqn. is obtained from general solution by giving particular values to all the arbitrary constants.
Hence the particular solution of a diff. eqn. of order 3 contains no arbitrary constants.

Question 3.
Verify that y = e^{– 3x} is a solution of the differential equation \(\frac{d y}{d x}\) + 3y = 0.
Solution:
Given y = e^{– 3x} ………………(1)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = – 3 e^{– 3x} = – 3y
⇒ \(\frac{d y}{d x}\) + 3y = 0, which is the given differential equaiton.
Hence y = e^{– 3x} be the soln. of given diff. eqn.

Question 3 (old).
What is the number of arbitrary constants in the general solution of the differential equation (1 + x²) (\(\frac{d^2 y}{d x^2}\))³ + y (\(\frac{d y}{d x}\))⁴ = 7 √x.
Solution:
Given diff. eqn. be,
(1 + x²) (\(\frac{d^2 y}{d x^2}\))³ + y (\(\frac{d y}{d x}\))⁴ = 7 √x
Since the order of given diff. eqn. be 2 as the highest order derivative existing in given diff. eqn. be \(\frac{d^2 y}{d x^2}\)
and its order be 2.
Thus the general solution of given diff. eqn. of order 2 contains 2 arbitrary constants.

Question 4.
In each of the following, show that the given function is a solution of the corresponding differential equation :
(i) y = cos x + C ; \(\frac{d y}{d x}\) + sin x = 0 (NCERT)
(ii) y = x² + 2x + C ; y’ – 2x – 2 = 0 (NCERT)
(iii) y = Ae^{– x} ; y’ + y = 0
(iv) y = Ax : xy’ = y (NCERT)
(v) y = e + 1 : y” – y’ = 0 (NCERT)
Solution:
(i) Given y = cos x + C …………..(1)
⇒ y’ = – sin x
⇒ y’ + sin x = 0 [Hence verified]

(ii) Given, y = x2 + 2x + c ………….(1)
Diff. eqn. (1) both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = 2x + 2
⇒ y’ – 2x – 2 = 0
which is the given diff. eqn.
Thus, y = x² + 2x + c be the solution of given differential eqn.

(iii) Given y = Ae^-x …………….(1)
Diff. eqn. (1) both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = – Ae^-x = – y [using eqn. (1)]
⇒ y’ + y = 0, which is given diff. eqn.
Hence, y = Ae^-x is the solution of given differential eqn.

(iv) Given y = Ax ……………(1)
Diff. eqn. (1) both sides w.r.t. x, we have
y’ = A
⇒ y’ = \(\frac{y}{x}\) [using eqn. (1)]
⇒ xy’ – y = 0, which is given differential eqn.
Thus, y = Ax be the solution of given diff. eqn.

(v) Given y = e^x + 1
∴ y’ = e^x
⇒ y” = e^x
⇒ y” = y’
⇒ y” – y’ = 0.

Question 5.
Show that the function y = ax + 2a² is a solution of the differential equation 2(\(\frac{d y}{d x}\))² + x \(\frac{d y}{d x}\) – y = 0.
Solution:
Given y = ax + 2a²
L.H.S = 2(\(\frac{d y}{d x}\))² + x \(\frac{d y}{d x}\) – y
= 2a² + ax – (ax + 2a²) = 0
= R.H.S.

Question 7 (old).
Show that y = c₁ e^x + c₂ e^{– x} is the general solution of the differential equation \(\frac{d^2 y}{d x^2}\) – y = 0.
Solution:
Given y = c₁ e^x + c₂ e^{– x} …………….(1)
Differentiating (1) w.r.t. x, we have
\(\frac{d y}{d x}\) = c₁ e^x + c₂ e^{– x} …………….(2)
diff. eqn. (2) w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = c₁ e^x + c₂ e^{– x} = y [using eqn. (1)]
⇒ \(\frac{d^2 y}{d x^2}\) – y = 0, which is given diff. eqn.
Thus, y = c₁ e^x + c₂ e^{– x} be the solution of the given diff. eqn.

In each of the following (8 to 14), show that the given function (explicit or implicit) is a solution of the corresponding differential equation :

Question 8.
y = \(\sqrt{a^2-x^2}\), x ∈(- a, a) : x + y \(\frac{d y}{d x}\) = 0. (NCERT)
Solution:
Given y= \(\sqrt{a^2-x^2}\) ;
on squaring
⇒ y² = a² – x²
⇒ x² + y² = a²
Diff. both sides w.r.t. x ; we have
2x + 2y \(\frac{d y}{d x}\) = 0
which is the same as given differential equation
Thus y = \(\sqrt{a^2-x^2}\) be the solution of x + y \(\frac{d y}{d x}\) = 0.

Question 9.
y² = 4ax : y = x \(\frac{d y}{d x}\) + a \(\frac{d x}{d y}\)
Solution:
Given y² = 4ax …………..(1).
Diff. eqn. (1) w.r.t. x ; we have
2y \(\frac{d y}{d x}\) = 4a
⇒ \(\frac{d y}{d x}=\frac{4 a}{2 y}=\frac{2 a}{y}\)
Now R.H.S.
⇒ \(x \frac{d y}{d x}+a \frac{d y}{d x}=\frac{2 a x}{y}+\frac{a y}{2 a}\)
= \(\frac{y^2}{2 y}+\frac{y}{2}\) [∵ Using (1)]
= y
= L.H.S
THus y² = 4ax be the solution of x \(\frac{d y}{d x}\) + a \(\frac{d x}{d y}\) = 1.

Question 9 (old).
y = \(\frac{1}{4}\) (x ± A)² : y₁² = y.
Solution:
Given y = \(\frac{1}{4}\) (x ± A)²
Diff. both sides w.r.t. x ; we have
\(\frac{d y}{d x}\) = \(\frac{1}{2}\) (x ± A)
On squaring both sides, we have
(\(\frac{d y}{d x}\))² = \(\frac{1}{4}\) (x ± A)² = y
Thus y = \(\frac{1}{4}\) (x ± A)² be the solution of given differential eqn (\(\frac{d y}{d x}\))² = y.

Question 10.
x + y = tan^-1 y : y²y’ + y^-1 + 1 = 0
Solution:
Given, x + y = tan^-1 y ……………(1)
Diff. eqn. (1) both sides w.r.t. x, we have
1 + y’ = \(\frac{1}{1+y^2}\) y’
⇒ (1 + y²) (1 + y’) = y’
⇒ 1 + y² + (1 + y² – 1) = y’
⇒ 1 + y² + y²y’ = 0 which is given diff. eqn.
Thus, x + y = tan^-1 y is the solution of given diff. eqn.

Question 11.
y = a cos x + b sin x ; \(\frac{d^2 y}{d x^2}\) + y = 0 (NCERT)
Solution:
Given y = a cos x + b sin x ………………(1)
Diff. both sides of eqn. (1) w.r.t. x ; we have
\(\frac{d y}{d x}\) = – a sin x + b cos x
again diff. w.r.t. x ; we have
\(\frac{d^2 y}{d x^2}\) = – a cos x – b sin x = – y
⇒ \(\frac{d^2 y}{d x^2}\) + y = 0 ……………..(2)

Question 12.
y = x sin 3x : \(\frac{d^2 y}{d x^2}\) + 9y – 6 cos 3x = 0 (NCERT)
Solution:
Given y = x sin 3x ……………….(1)
Diff. eqn. (1) both sides w.r.t. x ; we have
\(\frac{d y}{d x}\) = sin 3x + 3 x cos 3x ……………..(2)
Diff. eqn. (2) w.r.t. x ; we have
\(\frac{d^2 y}{d x^2}\) = 3 cos 3x + 3 (- 3x sin 3x + cos 3x)
⇒ \(\frac{d^2 y}{d x^2}\) = – 9x sin 3x + 6 cos 3x
⇒ \(\frac{d^2 y}{d x^2}\) + 9y – 6 cos 3x = 0 [using eqn. (1)]
Thus, y = x sin 3x be the solution of diff. eqn.

Question 13.
Show that y = e^-x + ax + b is a solution of the differential equation e^x \(\frac{d^2 y}{d x^2}\) = 1.
Solution:
Given y = e^-x + ax + b ………………..(1)
Diff. (1) w.r.t. x; we have
\(\frac{d y}{d x}\) = – e^-x + a ………….(2)
again diff. w.r.t x; we have
\(\frac{d^2 y}{d x^2}\) = + e^-x
e^x \(\frac{d^2 y}{d x^2}\) which is same as given differential equation.
Thus y = e^-x + ax + b be the solution of e^x \(\frac{d^2 y}{d x^2}\) = 1.

Question 13 (old).
y = e^{– 3x} : \(\frac{d^2 y}{d x^2}\) + \(\frac{d y}{d x}\) – 6y = 0
Solution:
Given y = e^{– 3x} ……………………(1)
\(\frac{d y}{d x}\) = – 3 e^{– 3x} ……………….(2)
Diff. eqn. (2) w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = 9 e^{– 3x} ………………(3)
Now, \(\frac{d^2 y}{d x^2}\) + \(\frac{d y}{d x}\) – 6y
= 9e^{– 3x} – 3e^{– 3x} – 6e^{– 3x} = 0
[using eqn. (1), (2) and (3)]
Hence y = e^{– 3x} be the solution of given differential eqn.

Question 14.
Show that y = a cos (log x) + b sin (log x) is a solution of the differential equation x² \(\frac{d^2 y}{d x^2}\) + x \(\frac{d y}{d x}\) + y = 0.
Solution:
Given y = a cos (log x) + b sin (log x) …………………(1)
Differentiating eqn. (1) w.r.t. x ; we have
\(\frac{d y}{d x}\) = – a sin (log x) . \(\frac{1}{x}\) + b cos (log x) . \(\frac{1}{x}\)
⇒ x \(\frac{d y}{d x}\) = – a sin (log x) + b cos (log x) …………………(2)
Differentiating eqn. (2) w.r.t. x ; we have
x \(\frac{d^2 y}{d x^2}\) + \(\frac{d y}{d x}\) . 1 = – a cos (log x) . \(\frac{1}{x}\) – b sin (log x) . \(\frac{1}{x2}\)
⇒ x² \(\frac{d^2 y}{d x^2}\) + x \(\frac{d y}{d x}\) = – y [using eqn. (1)]
⇒ x² \(\frac{d^2 y}{d x^2}\) + x \(\frac{d y}{d x}\) + y = 0, which is given diff. eqn.
Thus, y = a cos (log x) + b sin (log x) be the soln. of given diff. eqn.

Question 15.
Show that the differential equation of which y = 2 (x² – 1) + ce^{– x2} is a solution \(\frac{d y}{d x}\) + 2xy = 4x³.
Solution:
Given y = 2 (x² – 1) + ce^{– x2} ……………..(1)
Diff. both sides w.r.t. x; we have
∴ \(\frac{d y}{d x}\) = 4x + cce^{– x2} (- 2x)
\(\frac{d y}{d x}\) = 4x – 2x [y – 2x² + 2]
\(\frac{d y}{d x}\) = 4x – 2xy + 4x³ – 4x
\(\frac{d y}{d x}\) + 2xy = 4x³
which is same as given differential eqn.
Thus eqn. (1) be the solution of \(\frac{d y}{d x}\) + 2xy = 4x³..

Question 16.
Show that y² = 4a (x + a) is a solution of the differential equation y (1 – y₁²) = 2xy₁.
Solution:
Given y² = 4a (x + a) ……………….(1)
Diff. both sides of eqn (1) w.r.t. x; we have
2y \(\frac{d y}{d x}\) = 4a
⇒ \(\frac{d y}{d x}=\frac{2 a}{y}\) ……………..(2)

Thus y² = 4 a (x + a) be the solution of y \(\left\{1-\left(\frac{d y}{d x}\right)^2\right\}=2 x \frac{d y}{d x}\).

Question 17.
Show that y = sin (sin x) is a solution of the differential equation \(\frac{d^2 y}{d x^2}\) + tan x \(\frac{d y}{d x}\) + y cos² x = 0.
Solution:
Given, y = sin (sin x) …………………….(1)
Diff. eqn. (1) both sides w.r.t. x ; we have
\(\frac{d y}{d x}\) = cos (sin x) cos x ………………….(2)
Diff. eqn. (2) both sides w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = cos (sin x) (- sin x) + cos² x (- sin (sin x)) …………….(3)
From (2);
\(\frac{1}{\cos x} \frac{d y}{d x}\) = cos (sin x)
∴ from (3); we have
\(\frac{d^2 y}{d x^2}\) = – tan x \(\frac{d y}{d x}\) – y cos² x (using (1)]
⇒ \(\frac{d^2 y}{d x^2}\) + tan x \(\frac{d y}{d x}\) + y cos² x = 0
which is given diff. eqn.
Thus, y = sin (sin x) be the solution of given diff. eqn.

Question 18.
Show that y = e^{m sin-1 x} is a solution of the differential equation (1 – x²) y₂ – xy₁ – m²y = 0.
Solution:
Given y = e^{m sin-1 x} ……………………..(1)
Differentiating eqn. (1) w.r.t. x; we have
\(\frac{d y}{d x}\) = e^{m sin-1 x} \(\frac{m}{\sqrt{1-x^2}}\)
⇒ \(\sqrt{1-x^2} \frac{d y}{d x}\) = my ………………..(2) [using eqn. (1)]
Differentiating eqn. (2) w.r.t. x ; we have
\(\sqrt{1-x^2} \frac{d^2 y}{d x^2}+\frac{1}{2}\left(1-x^2\right)^{-\frac{1}{2}}(-2 x) \frac{d y}{d x}=m \frac{d y}{d x}\)
⇒ \(\sqrt{1-x^2} \frac{d^2 y}{d x^2}-\frac{x}{\sqrt{1-x^2}} \frac{d y}{d x}=m \sqrt{1-x^2} \frac{d y}{d x}\)
⇒ (1 – x²) y₂ – xy₁ = m²y [using eqn. (2)]
⇒ (1 – x²) y₂ – xy₁ – m²y = 0, which is given diff. eqn.
Thus, y = e^{m sin-1 x} be the soln. of given diff. eqn.