These NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.3 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.3
Question 1.
\(\frac{x}{a}+\frac{y}{b}\) = 1
Solution:
\(\frac{x}{a}+\frac{y}{b}\) = 1
i.e., bx + ay = ab … (1)
Differentiating (1) w.r.t. x, we get
b + ay’ = 0
i.e., ay’ = – b
y = \(\frac { – b }{ a }\)
Differentiating again w.r.t. x, we get y’ = 0
which is the required differential equation.
Question 2.
y² = a(b² – x²)
Solution:
y² = a(b² – x²)
i.e., y² = ab² – x²a
Differentiating w.r.t. x, we get 2yy’ = – 2ax
i.e., yy’ = – ax
\(\frac { yy’ }{ x }\) = – a
Differentiating again w.r.t. x, we get
\(\frac{x\left[y y^{\prime \prime}+\left(y^{\prime}\right)^{2}\right]-y y^{\prime}}{x^{2}}\) = 0
i.e., x yy” + x(y’)² – yy’ = 0
which is the required differential equation.
Question 3.
y = ae3x + be-2x
Solution:
Given that
y = ae3x + be-2x … (i)
Differentiating w.r.t. x two times, we get
y’ = 3ae3x – 2be-2x
y” = 9ae3x + 4be-2x
i.e., y” = 6ae3x + 3ae3x + 6be-2x – 2be-2x
= 3ae3x – 2be-2x + 6(ae3x + be-2x)
= y’ + 6y
i.e., y” – y’ – 6y = 0 is the required differential equation.
Question 4.
y = e2x (a + bx)
Solution:
y = e2x (a + bx)
y’ = be2x + (a + bx)e2x x 2
y’ = be2x + 2y
Differentiating again w.r.t. x, we get
y” = 2be2x + 2y’
i.e., y” = 2(y’ – 2y) + 2y’ [∵ y’ – 2y = be2x]
y” = 4y’ – 4y
y” – 4y’ + 4y = 0 is the required differential equation.
Question 5.
y = ex(a cosx + b sinx)
Solution:
The curve y = ex(a cosx+b sinx) …(i)
Differentiating (1) w.r.t. x, we get
y’ = ex(- a sinx + b cosx) + ex(a cosx + b sinx)
y’ = ex(b cos x – a sin y) + y
i.e., y’ – y = ex(b cosx – a sinx) … (2)
Differentiating (2) w.r.t. x, we get
y”- y’ = – ex(b sinx + a cosx) + ex(b cosx – a sinx)
y” – y’ = – y + y’- y [from (1) and (2)]
y” – 2y’ + 2y = 0 is the required differential equation.
Question 6.
Form the differential equation of the family of circles touching the y axis at origin
Solution:
Circles touching y axis at the orgin will have its centre on x-axis and will pass through the origin.
So the centre of circle will be (a, 0) and radius a.
∴ Equation of the circle is (x – a)² + (y – 0)² = a²
i.e., x² + y² – 2ax = 0
i.e., \(\frac{x^{2}+y^{2}}{x}\) = 2a
\(\frac{x\left(2 x+2 y y_{1}\right)-\left(x^{2}+y^{2}\right)}{x^{2}}\) = 0
Differentiating w.r.t. x, we get
2x² + 2xyy1 – x² – y² = 0
x² – y² + 2xyy1 = 0 or y² – x² – 2xyy1 = 0
Question 7.
Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
Solution:
Consider the family of parabolas having focus (0, a) at the positive y-axis, where a is an arbitrary constant.
∴ The equation of family of parabolas is x² = 4ay … (1)
Differentiating both sides w.r.t. x, we get
2x = 4 ay’
i.e., 4a = \(\frac { 2x }{ y’ }\) … (2)
Substituting (2) in (1) we get x² = (\(\frac { 2x }{ y’ }\))y
x²y’ – 2xy = 0
i.e., xy’ – 2y = 0 is the required differential equation.
Question 8.
Form the differential equation of family of ellipses having foci on y-axis and centre at origin.
Solution:
The equation of family ellipses having foci at y- axis is
Question 9.
Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.
Solution:
The equation of family of hyperbolas having foci on x axis is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1, a and b are the parameters.
i.e., xyy” + x(y’)² – yy’ = 0 is the required differential equation.
Question 10.
Form the differential equation of the family of circles having centre on y-axis and radius 3 units
Solution:
Consider a circle of radius 3 unit and centre on (0, a). The equation of the circle is
(x – 0)² + (y – a)² = 3²
x² + (y – a)² = 9 … (1)
Differentiating both sides w.r.t. x, we get
2x + 2(y – a)y’ = 0
x + yy’ – ay’ = 0
ay’ = x + yy’
⇒ (x² – 9)(y’)² + x² = 0 is the required differential equation.
Question 11.
Which of the following differential equation has y = \({ c }_{ 1 }{ e }^{ x }+{ c }_{ 2 }{ e }^{ -x }\) as the general solution?
(a) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +y=0\)
(b) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -y=0\)
(c) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +1=0\)
(d) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -1=0\)
Solution:
(b) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -y=0\)
Question 12.
Which of the following differential equations has y = x as one of its particular solution ?
(a) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -{ x }^{ 2 }\frac { dy }{ dx } +xy=x\)
(b) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ x }\frac { dy }{ dx } +xy=x\)
(c) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -{ x }^{ 2 }\frac { dy }{ dx } +xy=0\)
(d) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ x }\frac { dy }{ dx } +xy=0\)
Solution:
(c) y = x
