RD Sharma Solutions for Class 12 Maths Exercise 22.3 Chapter 22 Differential Equations assist students in learning complex and challenging topics. These solutions serve as an important study tool when it comes to preparing for the exam. Hence, students are suggested to solve all the questions provided in the RD Sharma Solutions for Class 12 Maths Chapter 22.
RD Sharma Solutions for Class 12 Maths Chapter 22 Exercise 3
Access RD Sharma Solutions for Class 12 Maths Chapter 22 Exercise 3
EXERCISE 22.3
Question. 1
Solution:
From the question, it is given that,
y = bex + ce2x … [equation (i)]
Now, differentiate the equation (i) with respect x,
dy/dx = bex + 2ce2x … [equation (ii)]
Then, the above equation is again differentiating with respect to x we get,
d2y/dx2 = bex + 4ce2x … [equation (iii)]
The given differential equation is d2y/dx2 – 3 (dy/dx) + 2y = 0
Substitute equation (i), equation (ii) and equation (iii) in the given differential equation,
d2y/dx2 – 3 (dy/dx) + 2y = 0
(bex + 4ce2x) – 3 (bex + 2ce2x) + 2(bex + ce2x) = 0
bex + 4ce2x – 3bex – 6ce2x + 2bex + 2ce2x = 0
3bex – 3bex + 6ce2x – 6ce2x = 0
0 = 0
Hence it is proved that, d2y/dx2 – 3 (dy/dx) + 2y = 0
Question. 2
Solution:
From the question, it is given that,
y = 4 sin 3x … [equation (i)]
Now, differentiate the equation (i) with respect to x,
dy/dx = 4(3) cos 3x
dy/dx = 12 cos 3x … [equation (ii)]
Then, the above equation is again differentiating with respect to x we get,
d2y/dx2 = 12(3) cos 3x
d2y/dx2 = – 36 sin 3x … [equation (iii)]
The given differential equation is d2y/dx2 + 9y = 0
Substitute equation (i) and equation (iii) in the given differential equation,
d2y/dx2 + 9y = 0
– 36 sin 3x + 9 (4 sin 3x) = 0
– 36 sin 3x + 36 sin 3x = 0
0 = 0
Hence it is verified that y = 4 sin 3x is a solution of the differential equation is d2y/dx2 + 9y = 0.
Question. 3
Solution:
From the question, it is given that,
y = ae2x + be-x … [equation (i)]
Now, differentiate the equation (i) with respect x,
dy/dx = 2 ae2x – be-x … [equation (ii)]
Then, the above equation is again differentiated with respect to x, and we get,
d2y/dx2 = 4 ae2x + be-x … [equation (iii)]
The given differential equation is d2y/dx2 – dy/dx – 2y = 0
Substitute equation (i), equation (ii) and equation (iii) in the given differential equation,
d2y/dx2 – dy/dx – 2y = 0
(4 ae2x + be-x) – (2 ae2x – be-x) – 2(ae2x + be-x) = 0
4 ae2x + be-x – 2ae2x + be-x – 2ae2x – 2be-x = 0
4 ae2x – 4ae2x + 2be-x – 2be-x = 0
0 = 0
Hence it is verified that y = ae2x + be-x is a solution of the differential equation is d2y/dx2 – dy/dx – 2y = 0.
Question. 4
Solution:
From the question, it is given that,
y = A cos x + B sin x … [equation (i)]
Now, differentiate the equation (i) with respect x,
dy/dx = – A sin x + B cos x … [equation (ii)]
Then, the above equation is again differentiating with respect to x we get,
d2y/dx2 = – A cos x – B sin x … [equation (iii)]
The given differential equation is d2y/dx2 + y = 0
Substitute equation (i) and equation (iii) in the given differential equation,
(- A cos x – B sin x) + (A cos x + B sin x ) = 0
– A cos x – B sin x + A cos x + B sin x = 0
0 = 0
Hence it is verified that y = A cos x + B sin x is a solution of the differential equation is d2y/dx2 + y = 0.
Question. 5
Solution:
From the question, it is given that,
y = A cos 2x – B sin 2x … [equation (i)]
Now, differentiate the equation (i) with respect x,
dy/dx = – 2A sin (2x) – 2B cos 2x
Taking common terms outside,
dy/dx = -2 (A sin 2x + B cos 2x) … [equation (ii)]
Then, the above equation is again differentiating with respect to x we get,
d2y/dx2 = – 2 [2A cos 2x – 2B sin 2x]
= -4 [A cos 2x – B sin 2x] … [equation (iii)]
The given differential equation is d2y/dx2 + 4y = 0
Substitute equation (i) and equation (iii) in the given differential equation,
d2y/dx2 + 4y = 0
-4 [A cos 2x – B sin 2x] + 4 (A cos 2x – B sin 2x) = 0
-4A cos 2x + 4B sin 2x + 4A cos 2x – 4B sin 2x = 0
0 = 0
Hence it is verified that y = A cos 2x – B sin 2x is a solution of the differential equation is d2y/dx2 + 4y = 0.
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