Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 9 Differential Equations Ex 9.3 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 12 Maths Chapter 9 Differential Equations Ex 9.3

In each of the questions 1 to 5, form a differential equation representing the given family of curves by eliminating arbitrary constants a and b:
Question 1.
(frac{x}{a}) + (frac{y}{b}) = 1
Solution:
(frac{x}{a}) + (frac{y}{b}) = 1
Differentiating w.r.t. x,
(frac{1}{a}) + (frac{y’}{b}) = 0.
Again differentiating,
(frac{1}{a}).y” = 0 or y” = 0.
∴ Required differential equation is y” = 0

Question 2.
y² = a(b² – x²)
Solution:
y² = a(b² – x²) ………………….. (1)
Differentiating w.r.t. x,
2yy’ = a(0 – 2x) = – 2ax ………………… (2)
Again differentiating,
2(y² + yy’) = – 2a ………………… (3)
Dividing (3) by (2) we get,

⇒ x(y² + yy”) = yy’,
which is the required differential equation.
It can also be written as xy(frac{d^{2} y}{d x^{2}}) + x((frac{dy}{dx}))² – y(frac{dy}{dx}) = 0.

Question 3.
y = a e^3x + b e^-2x
Solution:
y = a e^3x + b e^-2x ……………………. (1)
Differentiating w.r.t. x,
y’ = 3ae^3x – 2be^-2xx ……………….. (2)
Again differentiating,
y” = 9a^3x + 4be^-2x ………………….. (3)
Multiplying equation (1) by 2 and add with (2) to obtain:

Multiplying (1) by 3 and subtract (2) from it to obtain:

Putting values of ae^3x and be^-2x in (3), we get
y” = 9((frac{2y+y’}{5})) + 4((frac{3y-y’}{5}))
or 5y” = 9(2y + y’) + 4(3y – y’)
= 30y + 5y’
or y” – y’ – 6y = 0
∴ Required differential equation is
(frac{d^{2} y}{d x^{2}}) – (frac{dy}{dx}) – 6y = 0.

Question 4.
y = e^2x(a + bx)
Solution:
The curve is y = e^2x(a + bx) ………………. (1)
∴ y’ = 2e^2x(a + bx) + e^2x.
= e^2x(2a + b + 2bx) ……………….. (2)
Multiply eq. (1) and (2) and subtract it from (2) to get:

Differentiating eq. (3),
y” – 2y’ = 2be^2x …………….. (4)
Dividing (4) by (3), we get
or y” – 2y’ = 2(y’ – 2y)
or y” = 4y’ + 4y = 0
∴ The required differential equation is
(frac{d^{2} y}{d x^{2}}) – 4(frac{dy}{dx}) + 4y = 0.

Question 5.
y = e^x(a cos x + b sin x)
Solution:
The curve is y = e^x(a cos x + b sin x) ……………. (1)
Differentiating w.r.t. x, we get
y’ = e^x(a cos x + b sin x) + e^x(- a sin x + b cos x)
= e^x[(a + b)cos x – (a – b)sin x] ……………. (2)
Again differentiating, we get
y” = e^x(acos x + bsin x) + e^x(- asin x + bcos x)
= e^x[(a + b)cos x – (a – b)cos x]
= e^x[2b cos x – 2a sin x]
= 2e^x(b cos x – a sin x)
or (frac{y”}{2}) = e^x(b cos x – a sin x) ……………….. (3)
Adding (1) and (3),
y + (frac{y”}{2}) = e^x[(a + b)cos x – (a – b)sin x] = y’
or 2y + y” = 2y’
⇒ y” – 2y’ + 2y = 0.
Hence, the required differential equation is
(frac{d^{2} y}{d x^{2}}) – 2(frac{dy}{dx}) + 2y = 0.

Question 6.
From the differential equation of the family of circles touching the y-axis at origin.
Solution:
The equation of the circle with centre (a, 0) and radius a, which touches y-axis at origin, is
(x – a)² + y² = a²
or x² + y² = 2ax ……………. (1)

Differentiating w.r.t. x,
2x + 2y y’ = 2a
or x + yy’ = a
Putting value of a in (1), we get
x² + y² = 2x(x + yy’)
= 2x² + 2xy y’.
∴ Required differential equation is
2xy(frac{dy}{dx}) + x² – y² = 0.

Question 7.
Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
Solution:
Equation of parabola having vertex at the origin and axis along positive y-axis is
x² = 4ay ………….. (1)

where a is the parameter.
Differentiating w.r.t. x,
2x = 4ay’ ………………. (2)
Dividing (2) by (1) we get,
(frac{2 x}{x^{2}}) = (frac{2ay’}{4ay}) ⇒ (frac{y’}{y}) = (frac{2}{x})
∴ xy’ = 2y.
∴ Required differential equation is
x(frac{dy}{dx}) – 2y = 0.

Question 8.
Form the differential equation of family of ellipses having foci on y-axis.
Solution:
The equation of family of ellipses having foci on y-axis is
(frac{x^{2}}{b^{2}}) + (frac{y^{2}}{a^{2}}) = 1, a > b …………. (1)
Differentiating w.r.t. x,

Again differentiating,

Putting this value of (frac{1}{b^{2}}) in (2), we get

or x(y’² + yy”) = yy’
or xyy” + xy’² – yy’ = 0.
∴ Required differential equation is
xy (frac{d^{2} y}{d x^{2}}) + x((frac{dy}{dx}))² – y((frac{dy}{dx})) = 0.

Question 9.
Form the differential equation of the family of hyperbolas having foci on x-axis and the centre at origin.
Solution:
Equation on family of hyperbolas with centre at the origin and foci on x-axis is
(frac{x^{2}}{a^{2}}) – (frac{y^{2}}{b^{2}}) = 1 ………………. (1)
Differentiating w.r.t. x,

Again differentiating,

Question 10.
Form the differential equation of family of circles having centres on y-axis and radius 3 units.
Solution:
Let the centre be [0, b].
∴ Equation of family of circles of radius 3 is
x² + (y – b)² = 9 ………….. (1) [∵ r = 3]
Differentiating w.r.t. x,
2x + 2(y – b)y’ = 0 or y – b = – (frac{x}{y’})
Putting this value of (y – b) in (1), we get
x² + (- (frac{x}{y’}))² = 9
or x²y’² + x² = 9y’² or (x² – 9)y’² + x² = 0
∴ Required differential equation is (x² – 9) ((frac{dy}{dx}))² + x² = 0.

Choose the correct answers in the following questions 11 and 12:

Question 11.
Which of the following equations has y = c₁e^x + c₂e^-x as the general solution?
(A) (frac{d^{2} y}{d x^{2}}) + y = 0
(B) (frac{d^{2} y}{d x^{2}}) – y = 0
(C) (frac{d^{2} y}{d x^{2}}) + 1 = 0
(D) (frac{d^{2} y}{d x^{2}}) – 1 = 0
Solution:
Family of curve is y = c₁e^x + c₂e^-x …………….. (1)
Differentiating w.r.t. x,
y’ = c₁e_x – c₂e^-x
y” = c₁e^x + c₂e_-x = y
∴ y” – y = 0
Solution is (frac{d^{2} y}{d x^{2}}) – y = 0.
∴ Part (B) is the correct answer.

Question 12.
Which of the following differential equations has y = x as one of its particular solution?
(A) (frac{d^{2} y}{d x^{2}}) – x²(frac{dy}{dx}) + xy = 0
(B) (frac{d^{2} y}{d x^{2}}) + x(frac{dy}{dx}) + xy = x
(C) (frac{d^{2} y}{d x^{2}}) – x²(frac{dy}{dx}) + xy = 0
(D) (frac{d^{2} y}{d x^{2}}) + x (frac{dy}{dx}) + xy = 0
Solution:
The curve is y = x.
Differentiating w.r.t. x,
y’ = 1 and y” = 0.
Now consider part (C).
(frac{d^{2} y}{d x^{2}}) – x²(frac{dy}{dx}) + xy = 0 – x².1 + x . x
∵ y = x
= – x² + x² = 0.
Hence, Part (C) is the correct answer.