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 x^{7} = ln c

(v +1)^{5}. (v – 1)². x^{7} = c

(\(\frac{y}{x}\) + 1)^{5} (\(\frac{y}{x}\) – 1)².x^{7} = 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. X^{5}

= log(\(\frac{2}{(V+1)})\))².(\(\frac{3}{(V+2)}\))³. X^{5}

= log\(\frac{(Y+X)^2}{X^2}\) \(\frac{(Y+2X)^3}{X^3}\) . X^{5}

⇒ (Y + X)² . (Y + 2X)³ = e^{c} = 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