RBSE Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.5

Rajasthan Board RBSE Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.5 Textbook Exercise Questions and Answers.

RBSE Class 12 Maths Solutions Chapter 9 Differential Equations Ex 9.5

Question 1.
(x² + xy)dy = (x² + y²)dx
Answer:
Given differential equation is
(x² + xy)dy = (x² + y²)dx

Thus, f(x, y) is a homogeneous function of degree zero.
So, the given differential equation is also homogeneous.


Equation (5) is the required solution of the given differential equation.

Question 2.
y' = \(\frac{x+y}{x}\)
Answer:
Given differential equation is

Thus, f(x, y) is a homogeneous function of degree zero.
So, the given differential equation is also homogeneous.
Now putting y = vx ...... (2)
Then \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\) .......... (3)
From equations (1), (2) and (3)

Integrating both sides, we get
∫dv = ∫ \(\frac{d x}{x}\)
or v = log |x| + C
or \(\frac{y}{x}\) = log|x| + C
or y = x log |x| + Cx,
This is the required solution.

Question 3.
(x - y) dy - (x + y) dx = 0.
Answer:
Given differential equation is
(x - y)dy - (x + y)dx = 0

or f(λx, λy) = λ°f(x, y)
Thus, f(x, y) is a homogeneous function of degree zero.
So, the given differential equation is homogeneous.
Now putting y = vx ..... (2)

which is the required solution.

Question 4.
(x² - y²) dx + 2xy dy = 0
Answer:
Given differential equation is
(x² - y²) dx + 2xy dy = 0

= λ° f(x, y)
∴ f(x, y) is a homogeneous function of degree zero.
Thus, given differential equation is homogeneous.
Putting y = vx ........ (2)
Then \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\) ......... (3)
From equations (1), (2) and (3), we get

which is the required solution.

Question 5.
x²\(\frac{d y}{d x}\) = x² - 2y² + xy
Answer:
Given differential equation is

= λ° f(x, y)
∴ f(x, y) is a homogeneous function of degree zero.
Thus, the given differential equation is homogeneous.
Now, putting y = vx ....... (2)
Then \(\frac{d y}{d x} \)= v + x\(\frac{d v}{d x}\) .......... (3)
From equations (1), (2) and (3), we get

which is the required solution.

Question 6.
x dy - y dx = \(\sqrt{x^2+y^2}\) dx
Answer:
Given differential equation is
x dy - y dx = \(\sqrt{x^2+y^2}\) dx

or f(λx, λy) = λ⁰ f(x, y)
Thus f(x, y) is a homogeneous function of degree zero.
∴ So, the given differential equation is homogeneous.
Now, putting y = vx ............... (2)

which is the required solution.

Question 7.

Answer:
Given differential equation is

= λ⁰ f(x, y)
Thus f(x, y) is a homogeneous function of degree zero.
So, the given differential equation is homogeneous.
So putting y = vx ..... (2)
Then \(\frac{d y}{d x}\) = v + x\(\frac{d v}{d x}\) ......... (3)
From equations (1), (2) and (3), we get


which is the required solution.

Question 8.
x\(\frac{d y}{d x}\) - y + x sin\(\left(\frac{y}{x}\right) \)= 0
Answer:
Given differential equation is

Thus f(x, y) is a homogeneous function of degree zero.
∴ So, the given differential equation is homogeneous.
Putting y = vx ...... (2)

which is the required solution.

Question 9.
y dx + x log \(\left(\frac{y}{x}\right)\) dy - 2x dy = 0
Answer:
Given differential equation is

= λ⁰ f(x, y)
Thus, f(x, y) is a homogeneous function of degree zero.
∴ So, the given differential equation is homogeneous.
Putting y = vx ...... (2)

which is the required solution.

Question 10.
(1 + e^x/y) dx + e^x/y \(\left(1-\frac{x}{y}\right)\) dy = 0
Answer:
Given differential equation is

= λ⁰ f(x, y)
Thus, f(x, y) is a homogeneous function of degree zero.
So, the given differential equation is homogeneous.
Putting x = vy ............ (2)

This is the required solution of equation (1).

Question 11.
(x + y) dy + (x - y) dx = 0; y = 1 when x = 1.
Answer:
Given differential equation is
(x + y)dy + (x - y)dx = 0

Clearly, differential equation (1) is homogeneous.
Let, y = vx .......... (2)
Then \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\) ............ (3)
From equations (1), (2) and (3)

Let 1 + v² = t
or 2v dv = dt
∴ v dv = \(\frac{1}{2}\) dt
∫\(\frac{v d v}{1+v^2}\) = \(\frac{1}{2}∫\frac{d t}{t}\)
= \(\frac{1}{2}\) log |t|
= \(\frac{1}{2}\) log |1 + v²|

This is the required solution.

Question 12.
x² dy + (xy + y²) dx = 0; y = 1 when x = 1
Answer:
Given differential equation is
x² dy + (xy + y²) dx = 0

Clearly, it is homogeneous.
Let y = vx .......... (2)
Then \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\) ........... (3)
From equations (1), (2) and (3), we get

or x²y = C² (y + 2x)
Putting x = 1 and y = 1 in equation (4), we get
1² × 1 = C²(1 + 2 × 1)
or 1 = C² (3)
or C² = \(\frac{1}{3}\)
Putting the value of C² in equation (4), we get
x²y = \(\frac{1}{3}\)(y + 2x)
or y + 2x = 3x²y
which is the required solution.

Question 13.
[x sin² \(\left(\frac{y}{x}\right)\) - y] dx + x dy = 0; y = \(\frac{\pi}{4}\) when x = 1.
Answer:
Given differential equation is

Clearly, equation (1) is homogeneous differential equation.
Let y = vx .......... (2)
Then \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\) .............. (3)
From equations (1), (2) and (3), we get

Question 14.
\(\frac{d y}{d x}-\frac{y}{x}\) + cosec \(\left(\frac{y}{x}\right)\) = 0; y = 0 when x = 1.
Answer:
Given differential equation is
\(\frac{d y}{d x}-\frac{y}{x}\) + cosec \(\frac{y}{x}\) = 0
or \(\frac{d y}{d x}\) = \(\frac{y}{x}\) - cosec \(\frac{y}{x}\) ........... (1)
Clearly, equation (1) is, homogeneous differential equation.
Let y = vx .............. (2)
Then \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\) .............. (3)
From equations (1), (2) and (3), we get

which is the required solution.

Question 15.
2xy + y² - 2x²\(\frac{d y}{d x}\) = 0; y = 2 when x = 1
Answer:
Given differential equation is

Clearly, equation (1) is homogeneous differential equation
Let y = vx .............. (2)
Then \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\) .............. (3)
From equations (1), (2) and (3), we get

which is the required particular solution, when x ≠ 0, x ≠ e.

Question 16.
A homogeneous differential equation of the form \(\frac{d x}{d y}\) = h\(\left(\frac{x}{y}\right)\) can be solved by making the substitution:
(A) y = vx
(B) v = yx
(C) x = vy
(D) x = v
Answer:
When differential equation is the form of \(\frac{d x}{d y}\) = h\(\left(\frac{x}{y}\right)\), then we substitute x = vy.
Hence, option (C) is correct.

Question 17.
Which of the following is a homogeneous differential equation?
(A) (4x + 6y + 5) dy - (3y + 2x +4) dx = 0
(B) (xy)dx - (x³ + y³)dy = 0
(C) (x³ + 2y²)dx + 2xy dy = 0
(D) y² dx+ (x² - xy - y²)dy = 0.
Answer:
Differential equation y² dx + (x² - xy - y²) dy dt can be expressed as,

= λ⁰ f(x, y)
Thus f(x, y) is a homogeneous function of degree zero.
So, the given differential equation is homogeneous.
Hence, option (D) is correct.