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

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

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

I.
Question 1.
Express x dy – y dx = \(\sqrt{x^2+y^2}\) dx in the form \(F\left(\frac{y}{x}\right)=\frac{d y}{d x}\).
Solution:
Given equation is x dy – y dx = \(\sqrt{x^2+y^2}\) dx

Question 2.
Express (x – y tan^-1 \(\frac{y}{x}\)) dx + x tan^-1 \(\frac{y}{x}\) dy = 0 in the form \(F\left(\frac{y}{x}\right)=\frac{d y}{d x}\).
Solution:
Given equation is

Question 3.
Express x \(\frac{d y}{d x}\) = y (log y – log x + 1) in the form \(F\left(\frac{y}{x}\right)=\frac{d y}{d x}\).
Solution:
Given equation is x \(\frac{d y}{d x}\) = y (log y – log x + 1)
⇒ \(\frac{d y}{d x}=\frac{y}{x}\left[\log \left(\frac{y}{x}\right)+1\right]=F\left(\frac{y}{x}\right)\)

II. Solve the following differential equations.

Question 1.
\(\frac{d y}{d x}=\frac{x-y}{x+y}\)
Solution:
Let y = vx then \(\frac{d y}{d x}\) = v + x . \(\frac{d v}{d x}\)
∴ The given differential equation is

Question 2.
(x² + y²) dy = 2xy dx.
Solution:
The given differential equation is (x² + y²) dy = 2xy dx

∴ 1 + v² = A (1 – v²) + Bv (1 + v) + Cv (1 – v)
take v = – 1, we get
2C = – 2
⇒ C = – 1
coefficient of v² gives
⇒ – A + B – C = 1
⇒ – A + B = 0
coefficient of v gives B + C = 0
⇒ B = 1
∴ A = 1
∴ \(\frac{1+v^2}{v-v^3}=\frac{1}{v}+\frac{1}{1-v}-\frac{1}{1+v}\)
∴ From (1)
\(\int\left(\frac{1}{v}+\frac{1}{1-v}-\frac{1}{1+v}\right) \mathrm{d} v=\int \frac{\mathrm{dx}}{\mathrm{x}}+\log \mathrm{c}\)
⇒ log v – log (1 – v – log (1 + v) – log x + log c
⇒ log v – log(1 – v) – log (1 + v) = log cx
⇒ log v – [Iog(1 – v²)] = log cx

⇒ yx = cx (x² – y²)
⇒ y = c (x² – y²).

Question 3.
\(\frac{d y}{d x}=\frac{-\left(x^2+3 y^2\right)}{3 x^2+y^2}\)
Solution:

Question 4.
y² dx + (x² – xy) dy = 0
Solution:

⇒ u – log v = log x + log c
⇒ v = log (cx v)
⇒ \(\frac{y}{x}\) = log (cx \(\frac{y}{x}\)) = log (cy)
⇒ cy = e^y/x is the solution.

Question 5.
\(\frac{d y}{d x}=\frac{(x+y)^2}{2 x^2}\)
Solution:
Let y = vx
then \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)

2 tan^-1 v = log x + log c
2 tan^-1 (\(\frac{y}{x}\)) = log cx is the solution of the differential equation.

Question 6.
(x² – y²) dx – xy dy = 0
Solution:

Question 7.
(x²y – 2xy²) dx = (x³ – 3x²y) dy
Solution:

Question 8.
y² dx + (x² – xy + y²) dy = 0
Solution:

∴ 1 + v + v² = A (1 + v²) + (Bv + C) v
Comparing coefficient of v²,
A + B = 1
Also A = 1,
∴ B = 0.
Comparing coefficient of v,
C = – 1
∴ \(\frac{1-v+v^2}{v\left(1+v^2\right)}=\frac{1}{v}-\frac{1}{1+v^2}\)
∴ From (1)
\(\int \frac{1}{v} \mathrm{~d} v-\int \frac{1}{1+v^2} \mathrm{~d} v=-\int \frac{\mathrm{dx}}{\mathrm{x}}+\log \mathrm{c}\)
log v – tan^-1 v = – log x + log c

Question 9.
(y² – 2xy) dx + (2xy – x²) dy = 0
Solution:
The given equation is (y² – 2xy) dx = – (2xy – x²) dy

2v – 1 = A (1 – v) + 3Bv
Put v = 1,
1 = 3B
⇒ B = \(\frac{1}{3}\)
Also, – A + 3B = 2
⇒ 3B = 2 + A
⇒ 1 = 2 + A
⇒A = – 1
∴ \(\frac{2 v-1}{3 v(1-v)}=-\frac{1}{3 v}+\frac{1}{3} \frac{1}{1-v}\)
∴ From (1)

Question 10.
\(\frac{d y}{d x}+\frac{y}{x}=\frac{y^2}{x^2}\)
Solution:
Given \(\frac{d y}{d x}+\frac{y}{x}=\frac{y^2}{x^2}\)

v – 2 = c² x² . v
\(\frac{y}{x}\) – 2 = c²x² \(\frac{y}{x}\)
y – 2x = c²x³ \(\frac{y}{x}\)
= c²x²y
= kx²y
where c² = k
∴ Solution of the given equation is y – 2x= kx²y.

Question 11.
x dy – y dx = \(\sqrt{x^2+y^2}\) dx
Solution:
x dy = (y + \(\sqrt{x^2+y^2}\)) dx

Question 12.
(2x – y) dy = (2y – x) dx
Solution:
Given \(\frac{d y}{d x}=\frac{2 y-x}{2 x-y}\)
Let y = vx then
\(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)

⇒ (y – x)² = c² (y + x)² (y² – x²)
⇒ y – x = c² (y + x)³
⇒ (x + y)³ = c (x – y) where c = \(-\frac{1}{c^2}\) (constant)
Solution of the given differential equation is (x + y)³ = c (x – y)

Question 13.
(x² – y²) \(\frac{d y}{d x}\) = xy.
Solution:
The given equation is \(\frac{d y}{d x}\) = \(\frac{x y}{x^2-y^2}\)
Let y = vx then
\(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
v + x \(\frac{d v}{d x}\) = \(\frac{\mathrm{x}(v \mathrm{x})}{\mathrm{x}^2-v^2 \mathrm{x}^2}=\frac{v}{1-v^2}\)

∴ x² = – 2y² [log c + log y]
⇒ x² + 2y² (c + log y) = 0 is the solution of the given equation where log c = c.

Question 14.
2 \(\frac{d y}{d x}\) = \(\frac{y}{x}+\frac{y^2}{x^2}\)
then \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ 2 [v + x \(\frac{d v}{d x}\)] = v + v²
∴ 2v + 2x \(\frac{d v}{d x}\) = v + v²
⇒ 2x \(\frac{d v}{d x}\) = v + v² – 2v = v² – v
⇒ \(\frac{\mathrm{d} v}{v^2-v}=\frac{\mathrm{dx}}{2 \mathrm{x}}\) ………..(1)
writing in variable separable lorm
⇒ \(\frac{1}{v^2-v}=\frac{1}{v(v-1)}=\frac{\mathrm{A}}{v}+\frac{\mathrm{B}}{v-1}\)
∴ 1 = A (v – 1) + Bv
Put v = 1 then B = 1, and A + B = 0
⇒ A = – B = – 1
∴ \(\frac{1}{v^2-v}=-\frac{1}{v}+\frac{1}{v-1}\)
∴ From (1)

⇒ (y – x)² = y²xc²
⇒ c₁ (x – y)² = y²x
where c₁ = \(\frac{1}{c^2}\)
∴ The solution of the given differential equation is y²x = c₁ (x – y)².

III.
Question 1.
Solve: (1 + e^x/y) dx + e^x/y (1 – \(\frac{x}{y}\)) dy = 0
Solution:
The given equation is
(1 + e^x/y) dx + e^x/y (1 – \(\frac{x}{y}\)) dy = 0
⇒ (1 + e^x/y) \(\frac{d x}{d y}\) + e^x/y (1 – \(\frac{x}{y}\)) = 0
Let \(\frac{x}{y}\) = v then x = vy
∴ \(\frac{d x}{d y}\) = v + y \(\frac{d v}{d y}\)
∴ From (1)
(1 + e^y) (v + y \(\frac{d v}{d y}\) ) + e^v (1 – v) = 0

⇒ log (e^v + v) = – log y + log c
⇒ e^v + v = \(\frac{c}{y}\)
⇒ e^x/y + \(\frac{x}{y}\) = \(\frac{c}{y}\)
⇒ ye^x/y + x = c is the solution of the given equation.

Question 2.
Solve x sin \(\frac{y}{x}\) . \(\frac{d y}{d x}\) = y sin \(\frac{4}{4}\) – x.
Solution:

⇒ – cos v = – log x + log c
⇒ cos v = log x + log c = log (cx)
⇒ cos (\(\frac{y}{x}\)) = log (cx)
∴ The solution of the given equation is cx = e^cos(y/x).

Question 3.
x dy = (y + x cos² \(\frac{y}{x}\)) dx.
Solution:
Given x dy = (y + x cos² \(\frac{y}{x}\)) dx

⇒ ∫ sec² v dv = ∫ \(\frac{\mathrm{dx}}{\mathrm{x}}\) + c
⇒ tan v = log x + c
⇒ tan (\(\frac{y}{x}\)) = log x + c is the solution.

Question 4.
Solve (x – y log y + y log x) dx + x (log y – log x) dy = 0.
Solution:
Given (x – y log y + y log x) dx + x (log y – log x) dy = 0
⇒ x (log y – log x) dy = – (x – y log y + log x) dx

Writing in variable separable form we get
log v dv = – \(\frac{d x}{x}\)
∴ ∫ log v dv = – ∫ \(\frac{d x}{x}\) + c
⇒ v log v – v = – log x + c
⇒ \(\frac{y}{x} \log \left(\frac{y}{x}\right)-\frac{y}{x}\) = – log x + c
∴ Solution of the given differential equation is
\(\frac{y}{x} \log \left(\frac{y}{x}\right)-\frac{y}{x}\) = – log x + c
⇒ \(\frac{y}{x}\) (log y – log x) – \(\frac{y}{x}\) = – log x + c
⇒ y log y – y log x – y – x log x + cx
⇒ y log y + log x [(x – y)] = y + Cx.

Question 5.
Solve (y dx + x dy) x cos (\(\frac{y}{x}\)) = (x dy – y dx) y sin (\(\frac{y}{x}\))
Solution:
Given equation is
(y dx + x dy) x cos (\(\frac{y}{x}\)) = (x dy – y dx) y sin (\(\frac{y}{x}\))

Question 6.
Find the equation of a curve whose gradient is \(\frac{d y}{d x}=\frac{y}{x}-\cos ^2 \frac{y}{x}\) where x > 0, y > 0 and which passes through the point (1, \(\frac{\pi}{4}\)).
Solution:
Given gradient of the curve as \(\frac{d y}{d x}=\frac{y}{x}-\cos ^2 \frac{y}{x}\)
Let y = vx then
\(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ v + x \(\frac{d v}{d x}\) = v – cos² v
⇒ x \(\frac{d v}{d x}\) = – cos² v
⇒ x \(\frac{d v}{d x}\) = – cos² v
⇒ ∫ sec² v dv = – ∫ \(\frac{d x}{x}\)
⇒ tan v = – log x + log c
⇒ \(\tan \left(\frac{y}{x}\right)=\log \left(\frac{\mathrm{c}}{\mathrm{x}}\right)\)
Given curve passes through (1, \(\frac{\pi}{4}\)) we have
tan (\(\frac{\pi}{4}\)) = log (c)
⇒ c = e
∴ Solution of the given differential equation is
\(\tan \left(\frac{y}{x}\right)=\log \left(\frac{e}{x}\right)\)
= log e – log x
∴ Equation of the required curve is
tan \(\frac{y}{x}\) = 1 – log x.