Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Differential Equations Solutions Exercise 8(a) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2B Differential Equations Solutions Exercise 8(a)

I.

Question 1.

Find the order of the differential equation obtained by eliminating the arbitrary constants b and c from xy = ce^{x} – be^{-x} + x².

Solution:

Given equation is xy = ce^{x} – be^{-x} + x²

Differentiating w.r.to x, we get

xy_{1} + y = ce^{x} – be^{-x} + 2x.

Again differentiating w.r.to x, we get

xy_{2} + y_{1} + y_{1} = ce^{x} – be^{-x} + 2

xy_{2} + 2y_{2} = xy – x² + 2

Arbitary constants a and b are eliminated.

∴ The order is 2.

Question 2.

Find the order of the differential equation of the family of all circles with their centres at the origin.

Solution:

Equation of the circle with centre at origin is x² + y² = r²

Order = no .of arbitrary constants = 1

II.

Question 1.

Form the differential equations of the following family of curves where parameters are given in brackets.

i) y = c(x – c)² ; (c)

Solution:

y = c(x – c)² ………….. (1)

Differentiating w.r. to x

y_{1} = c. 2(x – c) ………….. (1)

Dividing (2) by (1)

\(\frac{y_2}{y}=\frac{2c(x-c)}{c(x-c)^2}\)

x – c = \(\frac{2y}{y_1}\)

c = x – \(\frac{2y}{y_1}\)

Substituting in (1)

y = x – \(\frac{2y}{y_1}\)(x – \(\frac{2y}{y_1}\))²

= \(\frac{xy_1-2y}{y_1}.\frac{4y^2}{y_1^2}\)

y.y³_{1} = 4y²(xy_{1} – 2y)

i.e., y³_{1} = 4y (xy_{1} – 2y)

= 4xyy_{1} – 8y²

(\(\frac{dy}{dx}\))³ – 4xy\(\frac{dy}{dx}\) + 8y² = 0

ii) xy = ae^{x} + be^{-x}; (a, b)

Solution:

xy = ae^{x} + b.e^{-x}

Differentiating w.r.t. x

x . y_{1} + y = ae^{x} – b . e^{-x}

Differentiating again w.r.t. x

xy_{2} + y_{1} + y_{1} = ae^{x} + be^{-x} = xy

\(\frac{d^2y}{dx^2}\) + 2\(\frac{dy}{dx}\) – xy = 0

iii) y = (a + bx)e^{kx} ; (a, b)

Solution:

y = (a + bx)e^{kx}

Differentiating w.r.t. x

y_{1} = (a + bx) e^{kx}. k + e^{kx} . b

= k . y + b.e^{kx}

y_{1} – ky = b.e^{kx} …………. (1)

Differentiating again w.r.t. x

y_{2} – ky_{1} = kb e^{kx}

= k(y_{1} – ky) ………… (2)

= ky_{1} – k²y

\(\frac{d^2y}{dx^2}\) – 2k\(\frac{dy}{dx}\) + k²y = 0

iv) y = a cos (nx + b); (a, b)

Solution:

y = a cos (nx + b)

y_{1} = – a sin (nx + b) n

y_{2} = – an. cos (nx + b) n

= – n² . y

\(\frac{d^2y}{dx^2}\)+n².y = 0

Question 2.

Obtain the differential equation which corresponds to each of the following family of curves.

i) The rectangular hyperbolas which have the co-ordinate axes as asymptotes.

Solution:

Equation of the rectangular hyperbolas is xy = c² where c is arbitrary constant

Differentiating w.r.t. x

x\(\frac{dy}{dx}\) + y = 0

ii) The ellipses with centres at the origin and having co-ordinate axes as axes.

Solution:

Equation of ellipse is

\(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1

Differentiating w.r.to ‘x’ we get

Multiply (ii) by x and subtract from (i)

III.

Question 1.

Form the differential equations of the following family of curves where parameters are given in brackets :

i) y = ae^{3x} + be^{3x}; (a, b)

Solution:

Differentiating w.r. to x

y_{1} = 3ae^{3x} + 4be^{4x}

y_{1} – 3a. e^{3x} = 4b.e^{4x}

= 4(y – a. e^{3x})

= 4y – 4a. e^{3x}

y_{1} – 4y = – a.e^{3x} ………… (1)

Differentiating again w.r.t. x

y_{2} – 4y_{1} = – 3a. e^{3x}

= 3 (y_{1} – 4y) by (1)

= 3y_{1} – 12y

\(\frac{d^2y}{dx^2}\) – 7\(\frac{dy}{dx}\) + 12y = 0

ii) y = ax² + bx; (a, b)

Solution:

Adding all three equations we get

x²\(\frac{d^2y}{dx^2}\) – 2x\(\frac{dy}{dx}\) + 2y = 0

iii) ax² + by² = 1; (a, b)

Solution:

ax² + by² = 1

by² = 1 – ax² ………….. (1)

Differentiating w.r.t. x

2by. y_{1} = – 2ax ………….. (2)

Dividing (2) by (1)

\(\frac{by.y_1}{by^2}=\frac{-ax}{1-ax^2}\)

Differentiating w.r.t. x

iv) xy = ax² + \(\frac{b}{x}\); (a, b)

Solution:

xy = ax² + \(\frac{b}{x}\)

x²y = ax³ + b

Differentiating w.r.t. x

x²y_{1} + 2xy = 3ax²

Dividing with x

xy_{1} + 2y = 3ax ………… (1)

Differentiating w.r.t. x

xy_{2} + y_{1} + 2y_{1} = 3a

xy_{2} + 3y_{1} = 3a ………… (2)

Dividing (1) by (2)

\(\frac{xy_1+2y}{xy_2+3y_1}=\frac{3ax}{3a}=x\)

Cross multiplying

xy_{1} + 2y = x²y_{2} + 3xy

x²y_{2} + 2xy_{1} – 2y = 0

x²\(\frac{d^2y}{dx^2}\) + 2x\(\frac{dy}{dx}\) – 2y = 0

Question 2.

Obtain the differential equation which corresponds to each of the following family of curves.

i) The circles which touch the Y – axis at the origin.

Solution:

Equation of the required circle is

x² + y² + 2gx = 0

x² + y² = – 2gx …………. (1)

Differentiating w.r. t x

2x + 2yy_{1} = – 2g ……….. (2)

Substituting in (1)

x² + y² = x(2x + 2yy_{1}) by (2)

= 2x² + 2xyy_{1}

yy² – 2xyy_{1} – 2x² = 0

y² – x² = 2xy\(\frac{dy}{dx}\)

ii) The parabolas each of which has a latus rectum 4a and whose axes are parallel to X – axis.

Solution:

Equation of the required parabola is

(y – k)² = 4a (x – h) …………. (1)

Differentiating w.r.t. x

2(y – k)y_{1} = 4a …………. (2)

Differentiating w.r.t. x

(y – k) y_{2} + y²_{1} = 0 …………. (3)

From (2), y – k = \(\frac{2a}{y_1}\)

Substituting in (3)

\(\frac{2a}{y_1}\).y_{2} = y²_{1} = 0

2ay_{2} + y³_{1} = 0

iii) The parabolas have their foci at the origin and axis along the X – axis.

Solution:

Equation of parabola be y² = 4a(x + a)