NCERT Solutions Class-12-Chapter-9-Diifferential Equations
Excercise-9.2
Note:
General Solution: Contains arbitrary constants equal to the order of the differential equation.
Particular Solution: Obtained from the general solution by giving specific values to arbitrary constants (contains no arbitrary constants).
To verify if a function is a solution, differentiate it and substitute the values of derivatives into the given differential equation to check if LHS = RHS.
Q1
Verify that the function is a solution of the differential equation:
\( y = e^x + 1 \) : \( y'' - y' = 0 \)
\( y = e^x + 1 \) : \( y'' - y' = 0 \)
Solution:
Given \( y = e^x + 1 \).
Differentiating w.r.t \( x \): \( y' = \frac{d}{dx}(e^x + 1) = e^x \).
Differentiating again: \( y'' = \frac{d}{dx}(e^x) = e^x \).
Substitute into the differential equation \( y'' - y' = 0 \):
LHS = \( e^x - e^x = 0 \) = RHS.
Hence, verified.
Q2
Verify that the function is a solution of the differential equation:
\( y = x^2 + 2x + C \) : \( y' - 2x - 2 = 0 \)
\( y = x^2 + 2x + C \) : \( y' - 2x - 2 = 0 \)
Solution:
Given \( y = x^2 + 2x + C \).
Differentiating w.r.t \( x \): \( y' = 2x + 2 \).
Substitute into the differential equation:
LHS = \( y' - 2x - 2 = (2x + 2) - 2x - 2 = 0 \) = RHS.
Hence, verified.
Q3
Verify that the function is a solution of the differential equation:
\( y = \cos x + C \) : \( y' + \sin x = 0 \)
\( y = \cos x + C \) : \( y' + \sin x = 0 \)
Solution:
Given \( y = \cos x + C \).
Differentiating w.r.t \( x \): \( y' = -\sin x \).
Substitute into the differential equation:
LHS = \( y' + \sin x = (-\sin x) + \sin x = 0 \) = RHS.
Hence, verified.
Q4
Verify that the function is a solution of the differential equation:
\( y = \sqrt{1+x^2} \) : \( y' = \frac{xy}{1+x^2} \)
\( y = \sqrt{1+x^2} \) : \( y' = \frac{xy}{1+x^2} \)
Solution:
Given \( y = \sqrt{1+x^2} \).
Differentiating w.r.t \( x \):
\( y' = \frac{1}{2\sqrt{1+x^2}} \cdot \frac{d}{dx}(1+x^2) = \frac{1}{2\sqrt{1+x^2}} \cdot 2x = \frac{x}{\sqrt{1+x^2}} \).
Now consider the RHS of the differential equation:
\( \frac{xy}{1+x^2} = \frac{x(\sqrt{1+x^2})}{1+x^2} = \frac{x}{\sqrt{1+x^2}} \).
Since LHS = RHS, it is verified.
Q5
Verify that the function is a solution of the differential equation:
\( y = Ax \) : \( xy' = y, (x \neq 0) \)
\( y = Ax \) : \( xy' = y, (x \neq 0) \)
Solution:
Given \( y = Ax \).
Differentiating: \( y' = A \).
Substitute into \( xy' = y \):
LHS = \( x(A) = Ax \).
RHS = \( y = Ax \).
LHS = RHS. Verified.
Q6
Verify that the function is a solution:
\( y = x \sin x \) : \( xy' = y + x\sqrt{x^2-y^2} \)
\( y = x \sin x \) : \( xy' = y + x\sqrt{x^2-y^2} \)
Solution:
Given \( y = x \sin x \).
Differentiating: \( y' = x \cos x + \sin x \).
LHS: \( xy' = x(x \cos x + \sin x) = x^2 \cos x + x \sin x \).
RHS: \( y + x\sqrt{x^2 - y^2} = x \sin x + x\sqrt{x^2 - x^2 \sin^2 x} \)
\( = x \sin x + x\sqrt{x^2(1 - \sin^2 x)} = x \sin x + x\sqrt{x^2 \cos^2 x} \)
\( = x \sin x + x(x \cos x) = x \sin x + x^2 \cos x \).
Since LHS = RHS, verified.
Q7
Verify that the function is a solution:
\( xy = \log y + C \) : \( y' = \frac{y^2}{1-xy} \)
\( xy = \log y + C \) : \( y' = \frac{y^2}{1-xy} \)
Solution:
Differentiating \( xy = \log y + C \) w.r.t \( x \):
\( x y' + y (1) = \frac{1}{y} y' \)
\( xy' - \frac{1}{y} y' = -y \)
\( y' (x - \frac{1}{y}) = -y \)
\( y' (\frac{xy-1}{y}) = -y \)
\( y' = \frac{-y^2}{xy-1} = \frac{y^2}{1-xy} \).
Verified.
Q8
Verify that the function is a solution:
\( y - \cos y = x \) : \( (y \sin y + \cos y + x)y' = y \)
\( y - \cos y = x \) : \( (y \sin y + \cos y + x)y' = y \)
Solution:
Differentiating \( y - \cos y = x \) w.r.t \( x \):
\( y' - (-\sin y) y' = 1 \)
\( y'(1 + \sin y) = 1 \Rightarrow y' = \frac{1}{1 + \sin y} \).
Substitute into LHS of differential equation:
LHS \( = (y \sin y + \cos y + x) \frac{1}{1+\sin y} \).
From the given function equation, \( x = y - \cos y \). Substitute \( x \):
\( = (y \sin y + \cos y + y - \cos y) \frac{1}{1+\sin y} \)
\( = (y \sin y + y) \frac{1}{1+\sin y} = \frac{y(\sin y + 1)}{1+\sin y} = y \).
LHS = y = RHS. Verified.
Q9
Verify that the function is a solution:
\( x + y = \tan^{-1} y \) : \( y^2 y' + y^2 + 1 = 0 \)
\( x + y = \tan^{-1} y \) : \( y^2 y' + y^2 + 1 = 0 \)
Solution:
Differentiating \( x + y = \tan^{-1} y \) w.r.t \( x \):
\( 1 + y' = \frac{1}{1+y^2} \cdot y' \)
\( 1 = y' \left( \frac{1}{1+y^2} - 1 \right) \)
\( 1 = y' \left( \frac{1 - (1+y^2)}{1+y^2} \right) = y' \left( \frac{-y^2}{1+y^2} \right) \)
\( 1+y^2 = -y^2 y' \)
\( y^2 y' + y^2 + 1 = 0 \).
Verified.
Q10
Verify that the function is a solution:
\( y = \sqrt{a^2 - x^2} \) : \( x + y \frac{dy}{dx} = 0 \)
\( y = \sqrt{a^2 - x^2} \) : \( x + y \frac{dy}{dx} = 0 \)
Solution:
Given \( y = \sqrt{a^2 - x^2} \). Squaring both sides: \( y^2 = a^2 - x^2 \Rightarrow x^2 + y^2 = a^2 \).
Differentiating w.r.t \( x \):
\( 2x + 2y \frac{dy}{dx} = 0 \)
Dividing by 2:
\( x + y \frac{dy}{dx} = 0 \).
Verified.
Q11
The number of arbitrary constants in the general solution of a differential equation of fourth order are:- 0
- 2
- 3
- 4
- 0
- 2
- 3
- 4
Answer: (D)
Solution:
The number of arbitrary constants in the general solution of a differential equation is equal to the order of the differential equation.
Since the order is 4, the number of arbitrary constants is 4.
Q12
The number of arbitrary constants in the particular solution of a differential equation of third order are:- 3
- 2
- 1
- 0
- 3
- 2
- 1
- 0
Answer: (D)
Solution:
A particular solution is obtained by assigning specific values to the arbitrary constants in the general solution. Therefore, a particular solution contains zero arbitrary constants, regardless of the order of the differential equation.