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 = cex – be-x + x².
Solution:
Given equation is xy = cex – be-x + x²
Differentiating w.r.to x, we get
xy1 + y = cex – be-x + 2x.
Again differentiating w.r.to x, we get
xy2 + y1 + y1 = cex – be-x + 2
xy2 + 2y2 = 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
y1 = 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.y³1 = 4y²(xy1 – 2y)
i.e., y³1 = 4y (xy1 – 2y)
= 4xyy1 – 8y²
($$\frac{dy}{dx}$$)³ – 4xy$$\frac{dy}{dx}$$ + 8y² = 0

ii) xy = aex + be-x; (a, b)
Solution:
xy = aex + b.e-x
Differentiating w.r.t. x
x . y1 + y = aex – b . e-x
Differentiating again w.r.t. x
xy2 + y1 + y1 = aex + be-x = xy
$$\frac{d^2y}{dx^2}$$ + 2$$\frac{dy}{dx}$$ – xy = 0

iii) y = (a + bx)ekx ; (a, b)
Solution:
y = (a + bx)ekx
Differentiating w.r.t. x
y1 = (a + bx) ekx. k + ekx . b
= k . y + b.ekx
y1 – ky = b.ekx …………. (1)
Differentiating again w.r.t. x
y2 – ky1 = kb ekx
= k(y1 – ky) ………… (2)
= ky1 – 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)
y1 = – a sin (nx + b) n
y2 = – 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 = ae3x + be3x; (a, b)
Solution:
Differentiating w.r. to x
y1 = 3ae3x + 4be4x
y1 – 3a. e3x = 4b.e4x
= 4(y – a. e3x)
= 4y – 4a. e3x
y1 – 4y = – a.e3x ………… (1)
Differentiating again w.r.t. x
y2 – 4y1 = – 3a. e3x
= 3 (y1 – 4y) by (1)
= 3y1 – 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. y1 = – 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
x²y1 + 2xy = 3ax²
Dividing with x
xy1 + 2y = 3ax ………… (1)
Differentiating w.r.t. x
xy2 + y1 + 2y1 = 3a
xy2 + 3y1 = 3a ………… (2)
Dividing (1) by (2)
$$\frac{xy_1+2y}{xy_2+3y_1}=\frac{3ax}{3a}=x$$
Cross multiplying
xy1 + 2y = x²y2 + 3xy
x²y2 + 2xy1 – 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 + 2yy1 = – 2g ……….. (2)
Substituting in (1)
x² + y² = x(2x + 2yy1) by (2)
= 2x² + 2xyy1
yy² – 2xyy1 – 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)y1 = 4a …………. (2)
Differentiating w.r.t. x
(y – k) y2 + y²1 = 0 …………. (3)
From (2), y – k = $$\frac{2a}{y_1}$$
Substituting in (3)
$$\frac{2a}{y_1}$$.y2 = y²1 = 0
2ay2 + y³1 = 0

iii) The parabolas having their foci at the origin and axis along the X – axis.
Solution:
Equation of parabola be y² = 4a(x + a)