Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d)

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.

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d)

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:
\(\frac{dy}{dx}=\frac{x+y+1}{2x+2y+3}\)
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 1
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:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 2
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

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d)

Question 3.
(2x + y + 1) dx + (4x + 2y – 1) dy = 0
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 3
∫(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:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 4
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:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 5
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 \(\frac{dy}{dx}=\frac{dy}{dx}\)
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 6
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 7
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:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 8
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 9
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 10
Multiplying with (3V – 2)(2V + 1)
2 + 18V = A(2V + 1) + B(3V – 2)
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 11
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
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 12
Solution is (3y – 2x – 1)² (x + 2y – 2) = 343c = c”.

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d)

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
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 13
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 14
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
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 15
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 16
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 17
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 18
is the required solution.

Question 5.
(x – y) dy = (x + y + 1) dx
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 19
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 20

Question 6.
(2x + 3y – 8) dx = (x + y – 3) dy
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 21
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 22
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 23
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 24

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d)

Question 7.
\(\frac{dy}{dx}=\frac{x+2y+3}{2x+3y+4}\)
Solution:
Let x = X + h, y = Y + k so that \(\frac{dY}{dX}=\frac{dy}{dx}\)
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 25
Choose h and k so that
h + 2k + 3 = 0
2h + 3k + 4 = 0
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 26
This is a homogeneous equation
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 27
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 28
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 29

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d)

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 \(\frac{dY}{dX}=\frac{dy}{dx}\)
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 30
∴ \(\frac{dY}{dX}=\frac{2X+9Y}{6X+2Y}\)
This is a homogeneous equation
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 31
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(d) 32