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

Students must practice this TS Intermediate Maths 2B Solutions Chapter 8 Differential Equations Ex 8(d) to find a better approach to solving the problems.

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

I.
Solve the following differential equations.

Question 1.
\(\frac{d y}{d x}=-\left(\frac{12 x+5 y-9}{5 x+2 y-4}\right)\)
Solution:
A non – homogeneous model of the type \(\frac{d y}{d x}=\frac{a x+b y+c}{a^{\prime} x+b^{\prime} y+c^{\prime}}\), where b = – a’
Here a = – 12, b = – 5, c = 9
a’ = 5, b’ = 2, c’ = – 4
So the solution of the differenüal equation can be obtained by Integrating each term alter regrouping
Given \(\frac{d y}{d x}=-\left(\frac{12 x+5 y-9}{5 x+2 y-4}\right)\)
∴ (5x + 2y – 4) dy = – (12x + 5y – 9) dx
∴ 5x dy + 2y dy – 4dy = – 12x dx – 5ydx + 9dx
⇒ 5 (x dy + y dx) + 2y dy – 4dy + 12x dx – 9 dx = 0
∴ 5 ∫ (x dy + y dx) + 2 ∫ y dy – 4 ∫ dy + 12 ∫ x dx – 9 ∫ dx = c
⇒ 5 ∫ d(xy) + 2 \(\frac{y^2}{2}\) – 4y + 12 \(\frac{x^2}{2}\) – 9x = c
⇒ 5xy + y² – 4y + 6x² – 9x = c
⇒ y² + 6x² + 5xy – 4y – 9x = c

Question 2.
\(\frac{d y}{d x}=\frac{-3 x-2 y+5}{2 x+3 y+5}\)
Solution:
Here a = – 3, b = – 2, c = 5
and a’ = 2, b’ = 3, c = 5
It is clear that b = – a’
∴ Solution is obtained by regrouping
2x dy + 3y dy + 5dy = – 3x dx – 2y dx + 5dx
⇒ 2(x dy + y dx) + 3(y dy + x dx) + 5dy – 5 dx = 0
Solution is
2 ∫ d(xy) + 3(\(\frac{y^2}{2}+\frac{x^2}{2}\)) + 5y – 5x = c
⇒ 2xy + \(\frac{3}{2}\) (x² + y²) + 5 (y – x) = c
⇒ 4xy + 3x² + 3y² + 10 (y – x) = 2c
⇒ 4xy + 3(x² + y²) – 10(x – y) = k.

Question 3.
\(\frac{d y}{d x}=\frac{-3 x-2 y+5}{2 x+3 y-5}\)
Solution:
Here a = – 3, b = – 2, c = 5
and a’ = – 2, b’ = 3, c = – 5
Here b = – a’ and hence solution can be obtained by regrouping
∴ 2x dy + 3y dy – 5dy = – 3x dx – 2y dx + 5dx
⇒ 2(x dy + y dx) + 3(y dy + x dx) – 5 dy – 5dx = 0
⇒ 2 ∫ d(xy) + 3 \(\left(\frac{y^2}{2}+\frac{x^2}{2}\right)\) – 5y – 5x = c
⇒ 2xy + 3 \(\left(\frac{y^2}{2}+\frac{x^2}{2}\right)\) – 5y – 5x = c
⇒ 4xy + 3(x² + y²) – 10y – 10x = k where k = 2c.

Question 4.
2 (x – 3y + 1) \(\frac{d y}{d x}\) = 4x – 2y + 1.
Solution:
\(\frac{d y}{d x}=\frac{4 x-2 y+1}{2 x-6 y+2}\)
Here a = 4, b = – 2, c = 1
and a’ = 2, b’ = – 6, c’ = 2
Here also b = – a’ and solution can be obtained by regrouping
∴ 2x dy – 6y dy + 2dy – 4x dx – 2y dx + dx
2 (x dy + y dx) – 6y dy – 4x dx + 2 dy – dx = 0
2 d(xy) – 6y dy – 4x dx + 2 dy – dx = 0
∴ 2 ∫ d(xy) – ∫ 6y dy – 4 ∫ x dx + 2 ∫ dy – ∫ dx = c
⇒ 2xy – 3y² – 2x² + 2y – x = c

Question 5.
\(\frac{d y}{d x}=\frac{x-y+2}{x+y-1}\)
Solution:
Here a = 1, b = – 1, c = 2
and a’ = 1, b’ = 1, c’ = – 1
Here b = – a’ and hence the solution can be obtained by regrouping of terms.
∴ x dy + y dy – dy = x dx – y dx + 2 dx
⇒ x dy + y dx + y dy – x dx – dy – 2dx = 0
⇒ ∫ d(xy) + ∫ y dy – ∫ x dx – ∫ dy – 2∫ dx = c
⇒ xy + \(\frac{y^2}{2}-\frac{x^2}{2}\) – y – 2x = c
⇒ 2xy + y² – x² – 2y – 4x = k where k = 2c
Solution is 2xy + y² – x² – 2y – 4x = k.

Question 6.
\(\frac{d y}{d x}=\frac{2 x-y+1}{x+2 y-3}\)
Solution:
Here a = 2, b = – 1, c = 1
and a’ = 1, b’ = 2, c’ = – 3
We have b = – a’ and the solution can be obtained by regrouping of terms.
∴ x dy + 2y dy – 3dy = 2x dx – y dx + dx
⇒ x dy + y dx + 2y dy – 2x dx – 3dy – dx = 0
⇒ d(xy) + 2y dy – 2x dx – 3 dy – dx = 0
∫ d(xy) + 2 ∫ y dy – 2 ∫ x dx – 3 ∫ dy – ∫ dx = c
⇒ xy + 2\(\left(\frac{y^2}{2}\right)\) 2 \(\left(\frac{x^2}{2}\right)\) – 3y – x = c
⇒ xy + y² – x² – 3y – x = c.

II.
Solve the following differential equations.
Question 1.
(2x + 2y + 3) \(\frac{d y}{d x}\) = x + y + 1
Solution:
\(\frac{d y}{d x}=\frac{x+y+1}{2 x+2 y+3}=\frac{x+y+1}{2(x+y)+3}\)
Let x + y = z then 1 + \(\frac{d y}{d x}\) = \(\frac{d z}{d x}\)

⇒ \(\frac{2}{3}\) z + \(\frac{1}{9}\) log (3z + 4) = x + c
⇒ \(\frac{2}{3}\) (x + y) + \(\frac{1}{9}\) log (3x + 3y + 4) = x + c
⇒ 6 (x + y) + log (3x + 3y + 4) = 9x + 9c
⇒ 6y – 3x + log (3x + 3y + 4) + c = 0 where c’ = – 9c.

Question 2.
\(\frac{d y}{d x}=\frac{4 x+6 y+5}{3 y+2 x+4}\)
Solution:

⇒ \(\frac{1}{8}\) (2x + 3y) + \(\frac{9}{64}\) log [8 (2x + 3y) + 23] = x + c
⇒ 8(2x + 3y) + \(\frac{9}{8}\) log(16x + 24y + 23) = 64x + 64c
⇒ 8[2x + 3y + log(16x + 24y + 23)] = 8 (8x + 8c)
⇒ 2x + 3y + \(\frac{9}{8}\) log (16x + 24y + 23) = 8x + 8c
⇒ 3y – 6x + \(\frac{9}{8}\) log (16x + 24y + 23) = k where k = 8c.

Question 3.
(2x + y + 1) dx + (4x + 2y – 1) dy = 0.
Solution:
From the given equation
\(\frac{d y}{d x}=-\left(\frac{2 x+y+1}{2(2 x+y)-1}\right)\)

⇒ \(\frac{2}{3}\) ∫ dz + \(\frac{1}{3} \int \frac{d z}{z-3}\) = x + c
⇒ \(\frac{2}{3}\) z + \(\frac{1}{3}\) log(z – 3) = x + c
⇒ \(\frac{2}{3}\) (2x + y) + \(\frac{1}{3}\) log(2x + y – 3) = x + c
⇒ (4x + 2y) + log (2x + y – 3) = 3x + 3c
⇒ (x + 2y) + log (2x + y + 3) = k where k = 3c

Question 4.
\(\frac{d y}{d x}=\frac{(2 y+x)+1}{2(x+2 y)+3}\)
Solution:
\(\frac{d y}{d x}=\frac{(2 y+x)+1}{2(x+2 y)+3}\)
Let x + 2y = z then 1 + 2 \(\frac{d y}{d x}\) = \(\frac{d z}{d x}\)

⇒ \(\frac{z}{2}\) + \(\frac{1}{8}\) log (4z + 5) = x + c
⇒ \(\frac{x+2 y}{2}\) + \(\frac{1}{8}\) [4(x + 2y) + 5] = x + c
⇒ 4x + 8y + log [4x + 8y + 5] = 8x + 8c
⇒ 8y – 4x + log (4x + 8y + 5) = 8k.

Question 5.
(x + y – 1) dy = (x + y + 1) dx
Solution:
\(\frac{d y}{d x}=\frac{x+y+1}{x+y-1}\)
Let x + y = z then 1 + \(\frac{d y}{d x}\) = \(\frac{d z}{d x}\)

⇒ \(\frac{1}{2}\) z – \(\frac{1}{2}\) log z = x + c
⇒ \(\frac{1}{2}\) (x + y) – \(\frac{1}{2}\) log (x + y) = x + c
⇒ x + y – log (x + y) = 2x + 2c
⇒ y – x – log (x + y) = k. where k = 2c.

III. Solve the following differential equations.

Question 1.
\(\frac{d y}{d x}=\frac{3 y-7 x+7}{3 x-7 y-3}\)
Solution:
\(\frac{d y}{d x}=-\left(\frac{7 x-3 y-7}{3 x-7 y-3}\right)\)
a = – 7, b = 3, c = 7
a’ = 3, b’ = – 7, c’ = – 3
b ≠ – a’ and \(\frac{a}{a^{\prime}} \neq \frac{b}{b^{\prime}}\)

Question 2.
\(\frac{d y}{d x}=\frac{6 x+5 y-7}{2 x+18 y-14}\)
Solution:
Let x = X + h and y = Y + k then
\(\frac{d y}{d x}=\frac{d Y}{d X}\)
and \(\frac{d Y}{d X}=\frac{6(X+h)+5(Y+k)-7}{2(X+h)+18(Y+k)-14}\)
= \(\frac{(6 \mathrm{X}+5 \mathrm{Y})+(6 \mathrm{~h}+5 \mathrm{k}-7)}{(2 \mathrm{X}+18 \mathrm{Y})+(2 \mathrm{~h}+18 \mathrm{k}-14)}\)
The equation becomes Non-homogeneous if
6h + 5k – 7 = 0
and 2h + 18k – 14 = 0
⇒ h + 9k – 7 = 0
Solving equations we get h = \(\frac{4}{7}\) and k = \(\frac{5}{7}\)

⇒ (2x – 3y + 1)² (x + 2y – 2) = k
⇒ (3y – 2x + 1)² (x + 2y – 2) = k is the solution of the given equation.

Question 3.
\(\frac{d y}{d x}+\frac{10 x+8 y-12}{7 x+5 y-9}\) = 0
Solution:
Given equation is \(\frac{d y}{d x}=-\frac{10 x+8 y-12}{7 x+5 y-9}\)

⇒ 3 log (v + 2) + 2 log (y + 1) = log X^-5 + log c
⇒ (v + 1)² (v + 2)³ X⁵ = c
⇒ (\(\frac{\mathrm{Y}}{\mathrm{X}}\) + 1)² (\(\frac{\mathrm{Y}}{\mathrm{X}}\) + 2)³ X⁵ = c
⇒ (Y + X)² (Y + 2X)³ = c
⇒ (y + 1 + x – 2)² [y + 1 + 2 (x – 2)]³ = c
⇒ (x + y – 1)² (2x + y – 3)³ = c is the solution of the equation.

Question 4.
(x – y – 2) dx + (x – 2y – 3) dy = 0
Solution:

Question 5.
(x – y) dy = (x + y + 1) dx
Solution:
The given equation can be written as \(\frac{d y}{d x}=\frac{x+\dot{y}+1}{x-y}\)
Let x = X + h, y = Y + k then

Question 6.
(2x + 3y – 8) dx = (x + y – 3)dy
Solution:
The given equation can be written as
\(\frac{d y}{d x}=\frac{2 x+3 y-8}{x+y-3}\)
Let x = X + h, y – Y + k then

Let 1 + v = A (2 – 2v) + B
then A = – \(\frac{1}{2}\) and 2A + B = 1
⇒ B = 1 – 2A
= 1 + 1 = 2
∴ 1 + v = – \(\frac{1}{2}\) (2 – 2v) + 2
∴ From (1)

Question 7.
\(\frac{d y}{d x}=\frac{x+2 y+3}{2 x+3 y+4}\)
Solution:
Let x = X + h, y = Y + k then
\(\frac{d Y}{d X}=\frac{X+h+2(Y+k)+3}{2(X+h)+3(Y+k)+4}\)
= \(\frac{X+2 Y+h+2 k+3}{2 X+3 Y+2 h+3 k+4}\)
Choose h and k such that h + 2k + 3 = 0 and 2h + 3k + 4 = 0
Solving we get h = 1, k = – 2

Question 8.
\(\frac{d y}{d x}=\frac{2 x+9 y-20}{6 x+2 y-10}\)
Solution:
Let x = X + h, y = Y + k then
\(\frac{\mathrm{dY}}{\mathrm{dX}}=\frac{2(\mathrm{X}+\mathrm{h})+9(\mathrm{Y}+\mathrm{k})-20}{6(\mathrm{X}+\mathrm{h})+2(\mathrm{Y}+\mathrm{k})-10}\)
= \(\frac{2 X+9 Y+(2 h+9 k-20)}{6 X+2 Y+(6 h+2 k-10)}\)
Choose h and k such that
2h + 9k – 20 = 0
and 6h + 2k – 10 = 0
Solving h = 1, k = 2

∴ 6 + 2v = A (1 + 2v) + B (2 – v)
Put v = 2 then 10 = 5A
⇒ A = 2
Also 2A – B = 2
⇒ – B = 2 – 2A = – 2
⇒ B = 2
∴ \(\frac{6+2 v}{2+3 v-2 v^2}=\frac{2}{2-v}+\frac{2}{1+2 v}\)
∴ From (1)
\(\int \frac{2}{2-v} \mathrm{~d} v+\int \frac{2}{1+2 v} \mathrm{~d} v=\int \frac{\mathrm{dX}}{\mathrm{X}}\)
⇒ – 2 log (2 – v) + log (1 + 2v) = log X + log c
⇒ log \(\frac{1}{(2-v)^2}\) + log (1 + 2v) = log cX
⇒ \(\frac{1}{(2-v)^2}\) . (1 + 2v) = cX
⇒ \(\frac{1}{\left(2-\frac{Y}{X}\right)^2}\left(1+\frac{2 Y}{X}\right)\) = cX
⇒ \(\frac{X^2}{(2 X-Y)^2}\left(\frac{X+2 Y}{X}\right)\) = cX
⇒ X + 2Y = c (2X – Y)²
⇒ [x – 1 + 2 (y – 2)] = c [2 (x – 1) – (y – 2)]²
⇒ (x + 2y – 5) = c [2x – y]²
⇒ (2x – y)² = \(\frac{1}{c}\) (x + 2y – 5) = c’ (x + 2y -5).