Class 12 NCERT Solutions- Mathematics Part II – Chapter 9 Differential Equations-Exercise -9.3
In each of the exercises 1 to 5, from a differential equation representing the given family of curves by eliminating arbitrary constraints a and b.
Question 1.
Solution:
Given:
We can also write
bx + ay = ab
On differentiating we get
b + ay’ = 0
y’ = -b/a
Again differentiating we get
y” = 0
Question 2.
Solution:
Given:
On differentiating we get
2y.y’=-2ax
Again differentiating we get
xyy” + x(y’)2 – yy’ = 0
Question 3. y = ae3x+be-2x
Solution:
y = ae3x + be-2x -(1)
On differentiating we get
y’=3ae3x-2be-2x -(2)
Again differentiating we get
y”=9ae3x+4be2x
Now on multiply eq(1) by 6
6y = 6ae3x + 6be-2x
And add with eq(2)
6y + y’ = 6ae3x + 6be-2x + 3ae3x – 3be-2x
6y + y’ = 9ae3x + 4be-2z = y”
y” – y’ – 6y = 0
Question 4. y = e2x(a + bx)
Solution:
Given: y = e2x(a + bx) -(1)
On differentiating we get
y’ = e2x(b) + (a + bx).2e2x
y’ = e2x(b + 2a + 2bx) -(2)
Now on multiply eq(1) by 2
2y = e2x(2a + 2bx)
And add with eq(2)
y’ – 2y = e2x(b + 2a + 2bx) – e2x(2a + 2bx)
y’ – 2y = be2x -(3)
Again differentiating we get
y” – 2y’ = 2be2x
Now put the value of be2x from eq(3)
y” – 2y’ = 2(y’ – 2y)
y” – 2y’ = 2y’ – 4y
y” – 2y’ – 2y’ + 4y = 0
y” – 4y’ + 4y = 0
Question 5.
Solution:
Given: y = ex(a cos x + b sin x) -(1)
On differentiating we get
y’ = ex[a cos x + b sin x – a sin x + b cos x]
y’ = y + ex[b cos x – a sin x] -(2)
Again differentiating we get
y’ ‘ =y’ + ex[b cos x – a sin x – b sin x – a cos x]
y” = y’ + ex[b cos x – a sin x] – ex[a cos x + b sin x]
From eq(1) and (2), we get
y” = y’ + [y’ – y] – y
y” – 2y’ + 2y = 0
Question 6. Form the differential equation of the family of circles touching the y-axis at the origin.
Solution:
Given that the family of circles touching the y-axis at the origin.
So, the center of the circle is (a, 0) and radius a
Let the equation of a circle is
(x – a)2 + y2 = a2
= x2 + y2 = 2ax -(1)
On differentiating we get
2x + 2yy’ = 2a
x + yy’ = a
Now substitute the value of a in eq(1), we get
x2 + y2 = 2(x + yy’)x
x2 + y2 = 2x2 + 2xyy’
x2 + y2 – 2x2 – 2xyy’
y2 = x2 + 2xyy’
Question 7. Form the differential equation of the family of parabolas having a vertex at origin and axis along positive y-axis.
Solution:
Given that the family of parabolas having a vertex at origin and axis along positive y-axis.
So the equation of parabola is:
x2 = 4ay -(1)
On differentiating we get
2x = 4ay’ -(2)
Now divide eq(2) by (1), we have
2x/ x2 = 4ay’ /4ay
2/x = y’ /y
y’x = 2y
y’x – 2y = 0
Question 8. Form the differential equation of the family of ellipses having foci on y-axis and center at the origin.
Solution:
Given that the family of ellipses having foci on y-axis & center at the origin.
So the equation of parabola is
On differentiating we get
Again differentiating we get
xyy” + x(y’)2 – yy’ = 0
Question 9. Form the differential equation of the family of hyperbolas having foci on the x-axis and center at the origin.
Solution:
Given that the family of hyperbolas having foci on the x-axis and center at the origin.
So the equation of hyperbola is
On differentiating we get
Again differentiating we get
Question 10. Form the differential equation of the family of circles having a center on y-axis and radius 3 units.
Solution:
Given that the of circles having a center on y-axis and radius 3 units.
So the center be (0, k)
General equation of the circle is,
x2 + (y – k)2 = 32
(y – k)2 = 9 – x2
y – k =
k =
On differentiating we get
Squaring on both sides we get
Question 11. Which of the following differential equations has y = c1ex + c2e-x as the general solution?
(A)
(B)
(C)
(D)
Solution:
y = c1ex + c2e-x
On differentiating we get
y’ = c1ex – c2e-x
Again differentiating we get
y” = c1ex + c2e-x
y” = y
y” – y = 0
Hence, the correct option is B
Question 12. Which of the following differential equations has y = x as one of its particular solution?
(A)
(B)
(C)
(D)
Solution:
y = x
On differentiating we get
y’ = 1
Again differentiating we get
y” = 0
Now substitute the value of y, y’ and y” in each option to check for correct option
(A)
= 0 – x2(1) + x.x = x
0 ≠x
(B)
= 0 + x(1) + x.x = x
= x + x2≠x
(C)
= 0 – x2(1) + x.x = 0
= 0 = 0
Hence, the correct option is C.
Last Updated :
31 Oct, 2022
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