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

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

I. Solve the following differential equations.

Question 1.
$$\frac{dy}{dx}=-\frac{(12x+5y-9)}{5x+2y-4}$$
Solution:
A non-homogenous equation
$$\frac{dy}{dx}=-\frac{(ax+by-9)}{a’x+b’y-c’}$$ where b = -a’
b = -5, a = 5 ⇒ b = -a
(5x + 2y-4)dy = -(12x + 5y-9) dx
(5x + 2y – 4)dy + (12x + 5y – 9) dx = 0
5 (x dy + y dx) + 2y dy – 4 dy + 12x dx – 9 dx = 0
integrating 5xy + y² – 4y + 6x² – 9x = c

Question 2.
$$\frac{dy}{dx}=-\frac{-3x-2y+5}{2x+3y+5}$$
Solution:
b = – 2, a = 2 ⇒ b = -a
(2x + 3y + 5) dy = (- 3x – 2y + 5) dx
2x dy + 3y dy + 5 dy = -3x dx- 2y dx + 5 dx
2(x.dy + y dx) + By dy + 3x dx + 5 dy – 5 dx = 0
Integrating
2xy + $$\frac{3}{2}$$y² + $$\frac{3}{2}$$x² + 5y – 5x = c
4xy + 3y² + 3x² – 10x + 10y = 2c = c’
Solution is
4xy + 3(x² + y²)- 10(x – y) = c

Question 3.
$$\frac{dy}{dx}=\frac{-3x-2y+5}{2x+3y-5}$$
Solution:
$$\frac{dy}{dx}=\frac{-(3x-2y+5)}{2x+3y-5}$$
Here b = – 2, a¹ = 2
∵ b = -a¹
(2x + 3y – 5) dy = (-3x – 2y + 5) dx „
⇒ 2(x dy + y dx) + (3y – 5) dy + (3x – 5) dx – 0
⇒ 2d (xy) + (3y- 5) dy + (3x- 5) dx = 0
Now integrating term by term, we get
⇒ 2 ∫d (xy) + ∫(3y – 5)dy + ∫(3x – 5)dx = 0
⇒ 2xy + 3.$$\frac{y^2}{2}$$ – 5y + 3$$\frac{x^2}{2}$$ – 5x = $$\frac{c}{2}$$
or) 3x² + 4xy + 3y² – 10x – 10y = c
Which is the required solution. Question 4.
2(x – 3y + 1) $$\frac{dy}{dx}$$ = 4x – 2y + 1
Solution:
(2x – 6y + 2) dy = (4x – 2y + 1) dx
(2x – 6y + 2) dy – (4x – 2y + 1) dx = 0
2 (x dy + y dx) – 6y dy + 2 dy – 4x dx – dx = 0
Integrating
2xy – 3y² – 2x² + 2y – x = c

Question 5.
$$\frac{dy}{dx}=\frac{x-y+2}{x+y-1}$$
Solution:
b = -1, a’ = 1 ⇒ b = -a’
(x + y – 1) dy = (x – y + 2) dx
(x + y – 1) dy = (x – y + 2) dx = 0
(x dy + y dx) + y dy – dy – x dx – 2 dx = 0
integrating
xy + $$\frac{y^2}{2}$$ – $$\frac{x^2}{2}$$ – y – 2x = c
2xy + y² – x² – 2y – 4x = 2c = c’

Question 6.
$$\frac{dy}{dx}=\frac{2x-y+1}{x+2y-3}$$
Solution:
b = -1, a = 1 ⇒ b = -a’
(x + 2y – 3) dy = (2x – y + 1) dx
(x + 2y – 3) dy – (2x – y + 1) dx = 0
(x dy + y dx) 4- 2y dy – 3 dy – 2x dx – dx = 0
Integrating
xy + y² – x² – 3y – x = c

II. Solve the following differential equations.

Question 1.
(2x + 2y + 3) $$\frac{dy}{dx}$$ = x + y + 1
Solution:
$$\frac{dy}{dx}=\frac{x+y+1}{2x+2y+3}$$ Multiplying with 9
6v + log (3v + 4) = 9x + 9c
6(x + y) + log [3(x + y) + 4] = 9x + c
i.e., log (3x + 3y + 4) = 3x – 6y + c

Question 2.
$$\frac{dy}{dx}=\frac{4x+6y+5}{3y+2x+4}$$
Solution: Multiplying with 64
8v + 9log (8v + 23) = 64x + 64c
8 (2x + 3y) – 64x + 9 log (16x + 24y + 23) = c’
Dividing with 8
2x + 3y – 8x + $$\frac{9}{8}$$ log (16x + 24y + 23) = c”
3y – 6x + $$\frac{9}{8}$$ log (16x + 24y + 23) = c”
Dividing with 3, solution is 3
y – 2x + $$\frac{3}{8}$$ log (16x + 24y + 23) = k Question 3.
(2x + y + 1) dx + (4x + 2y – 1) dy = 0
Solution: ∫(2 + $$\frac{1}{v-1}$$)dv = 3∫dx
2v + log (v – 1) = 3x + c
2v – 3x + log (v – 1) = c
2(2x + y) – 3x + log (2x + y – 1) = c
4x + 2y – 3x + log (2x + y – 1) = c
Solution is x + 2y + log (2x + y – 1) = c

Question 4.
$$\frac{dy}{dx}=\frac{2y+x+1}{2x+4y+3}$$
Solution: Multiplying with 8
4v + log (4v + 5) = 8x + 8c
4(x + 2y) – 8x + log [4(x + 2y) + 5] = c’
Solution is
4x + 8y – 8x + log (4x + 8y + 5) = c’
8y – 4x + log (4x + 8y + 5) = c’

Question 5.
(x + y – 1) dy = (x + y + 1)dx
Solution: v – log v = 2x + c
x + y – log (x + y) = 2x – c
(x – y) + log (x + y) = c is the required
solution.

III. Solve the following differential equations.

Question 1.
$$\frac{dy}{dx}=\frac{3y-7x+7}{3x-7y-3}$$
Solution:
Let x = x + h, y = y + k so that $$\frac{dy}{dx}=\frac{dy}{dx}$$  3ln (v – 1) – 3ln (v + 1) – 7ln (v + 1) – 7ln (v – 1)
14ln x – ln c = – 10ln (v + 1) – 4 ln (v – 1)
ln (v + 1)5 + ln (v – 1)² + ln x7 = ln c
(v +1)5. (v – 1)². x7 = c
($$\frac{y}{x}$$ + 1)5 ($$\frac{y}{x}$$ – 1)².x7 = c
(y – x)² (y + x)5 = c
[y – (x – 1 )]² (y + x – 1 )5 = c
Solution is [y-x + 1 ]² (y + x – 1)5 = c.

Question 2.
$$\frac{dy}{dx}=\frac{6x+5y-7}{2x+18y-14}$$
Solution:   Multiplying with (3V – 2)(2V + 1)
2 + 18V = A(2V + 1) + B(3V – 2) 2 log (3V- 2)+ log (2V+ 1) = – 3 log X + log c
log (3V – 2)².(2V + 1) + log X³ = log c
log X³(3V – 2)² (2V + 1) = log c
x³(3V – 2)² (2V + 1) = c Solution is (3y – 2x – 1)² (x + 2y – 2) = 343c = c”. Question 3.
$$\frac{dy}{dx}=\frac{10x+8y-12}{7x+5y-9}$$ = 0
Solution:
$$\frac{dy}{dx}=\frac{10x+8y-12}{7x+5y-9}$$ = 0
x = X + h, y = Y + k  5V + 7 = A(V + 2) + B (V + 1)
V = -1 ⇒ 2 = A(-1 + 2) = A ⇒ A = 2
V = -2 ⇒ -3 = B(-2 + 1) = -B, B = 3
∫($$\frac{2}{(V+1)}+\frac{3}{(V+2)}$$)dv = ∫$$\frac{dx}{X}$$
2 log (V + 1) + 3 log (V + 2) = – 5 log X + c
c = 2 log (V + 1) + 3 log (V + 2) + 5 log X
= log (V + 1)². (V + 2)³. X5
= log($$\frac{2}{(V+1)})$$)².($$\frac{3}{(V+2)}$$)³. X5
= log$$\frac{(Y+X)^2}{X^2}$$ $$\frac{(Y+2X)^3}{X^3}$$ . X5
⇒ (Y + X)² . (Y + 2X)³ = ec = c’
(Y + 1 – X – 2)² (Y + 1 – 2x – 4)³ = c
Solution is (x + y – 1)² (2x + y – 3)³ = c.

Question 4.
(x – y – 2) dx + (x – 2y – 3) dy = 0
Solution:
Given equation is $$\frac{dy}{dx}=\frac{-x+y+2}{x-2y-3}$$
Let x = X + h, y = Y + k    is the required solution.

Question 5.
(x – y) dy = (x + y + 1) dx
Solution:  Question 6.
(2x + 3y – 8) dx = (x + y – 3) dy
Solution:     Question 7.
$$\frac{dy}{dx}=\frac{x+2y+3}{2x+3y+4}$$
Solution:
Let x = X + h, y = Y + k so that $$\frac{dY}{dX}=\frac{dy}{dx}$$ Choose h and k so that
h + 2k + 3 = 0
2h + 3k + 4 = 0 This is a homogeneous equation    Question 8.
$$\frac{dy}{dx}=\frac{2x+9y-20}{6x+2y-10}$$
Solution:
Given equation is $$\frac{dy}{dx}=\frac{2x+9y-20}{6x+2y-10}$$
Let x = X + h, y = Y + k so that $$\frac{dY}{dX}=\frac{dy}{dx}$$ ∴ $$\frac{dY}{dX}=\frac{2X+9Y}{6X+2Y}$$
This is a homogeneous equation  