Q1: Verify that the given function y = e^{x} + 1 is a solution of the differential equation y'' - y' = 0 .
Solution:
Step 1: Write down the given function
Given function: y = e^{x} + 1 … (1)
Step 2: Find the First Derivative (y')
Differentiating equation (1) with respect to x :
y' = \frac{d}{dx} (e^x+1)
y' = e^{x} … (2)
Step 3: Find the Second Derivative (y'')
We differentiate equation (2) with respect to x :
y'' = \frac{d}{dx}(e^x)
y'' = e^{x} … (3)
Step 4: Substitute values into the Differential Equation
Substituting the values of y' and y'' into the Left Hand Side (L.H.S.) of the given differential equation:
\text{L.H.S } = y'' - y'
\text{L.H.S. } = e^{x} - e^{x}
\text{L.H.S.} = 0
Step 5: Compare with Right Hand Side
The Right Hand Side (R.H.S.) of the given equation is 0.
\text{L.H.S. } = \text{R.H.S. }
Conclusion: Since the L.H.S. equals the R.H.S., the function y = e^{x} + 1 satisfies the given differential equation.
Thus, it is a solution.
Q2: Verify that the given function y = x^{2} + 2x + C is a solution of the differential equation y' - 2x - 2 = 0 .
Solution:
Step 1: Write down the given function
Given function: y = x^{2} + 2x + C … (1)
Step 2: Find the First Derivative (y')
Differentiating equation (1) with respect to x :
y' = \frac{d}{dx} (x^2 + 2x +C)
y' = 2x + 2 … (2)
Step 3: Substitute values into the Differential Equation
Substituting the value of y' from equation (2) into the Left Hand Side (L.H.S.) of the given differential equation:
\text{L.H.S. } = y' - 2x -2
\text{L.H.S. } = (2x + 2) - 2x - 2
\text{L.H.S. } = 0
Step 4: Compare with Right Hand Side
The Right Hand Side (R.H.S.) of the given equation is 0.
\text{L.H.S. } = \text{R.H.S }
Conclusion: Since the L.H.S. equals the R.H.S., the function y = x^{2} + 2x + C satisfies the given differential equation .
Thus, it is a solution.
Q3: Verify that the given function y = \cos x + C is a solution of the differential equation y' + \sin x = 0 .
Solution:
Step 1: Write down the given function
Given function: y = \cos x + C … (1)
Step 2: Find the First Derivative (y')
Differentiating equation (1) with respect to x :
y' = \frac{d}{dx} (\cos x + C)
y' = -\sin x … (2)
Step 3: Substitute values into the Differential Equation
Substituting the value of y' from equation (2) into the Left Hand Side (L.H.S.) of the given differential equation:
\text{L.H.S } = y' + \sin x
\text{L.H.S. } = (-\sin x) + \sin x
\text{L.H.S. } = 0
Step 4: Compare with Right Hand Side
The Right Hand Side (R.H.S.) of the given equation is 0.
\text{L.H.S } = \text{R.H.S. }
Conclusion: Since the L.H.S. equals the R.H.S., the function y = \cos x + C satisfies the given differential equation .
Thus, it is a solution.
Q4: Verify that the given function y = \sqrt{1+x^2} is a solution of the differential equation y' = \frac{xy}{1+x^2} .
Solution:
Step 1: Write down the given function
Given function: y = \sqrt{1+x^2} … (1)
Step 2: Find the First Derivative (y')
Differentiating equation (1) with respect to x using the chain rule:
y' = \frac{d}{dx} (1+x^2)^{\frac{1}{2}}
y' = \frac{1}{2}(1+x^2)^{-\frac{1}{2}} \cdot \frac{d}{dx}(1+x^2)
y' = \frac{1}{2\sqrt{1+x^2}} \cdot (2x)
y' = \frac{x}{\sqrt{1+x^2}} … (2)
Step 3: Simplify the Right Hand Side (R.H.S.) of the Differential Equation
The given differential equation is y' = \frac{xy}{1+x^2} .
Substituting the value of y from equation (1) into the R.H.S.:
\text{R.H.S } = \frac{x(\sqrt{1+x^2})}{1+x^2}
\text{R.H.S } = \frac{x}{\sqrt{1+x^2}}
Step 4: Compare L.H.S. and R.H.S.
From Step 2, we have \text{L.H.S. } (y') = \frac{x}{\sqrt{1+x^2}} .
From Step 3, we have \text{R.H.S. } = \frac{x}{\sqrt{1+x^2}} .
\text{L.H.S. } = \text{R.H.S. }
Conclusion: Since the derivative matches the expression given in the differential equation, the function y = \sqrt{1+x^2} satisfies the given differential equation .
Thus, it is a solution.
Q5: Verify that the given function y = Ax is a solution of the differential equation xy' = y , (x \ne 0) .
Solution:
Step 1: Write down the given function
Given function: y = Ax … (1)
Step 2: Find the First Derivative (y')
Differentiating equation (1) with respect to x :
y' = \frac{d}{dx} (Ax)
y' = A … (2)
Step 3: Substitute values into the Differential Equation
We substitute the values of y' and y into the equation xy' = y .
Left Hand Side (L.H.S.):
\text{L.H.S. } = x(y')
Substituting y' = A :
\text{L.H.S. } = Ax
Right Hand Side (R.H.S.):
\text{R.H.S. } = y
Substituting y = Ax :
\text{R.H.S. } = Ax
Step 4: Compare L.H.S. and R.H.S.
\text{L.H.S. } = \text{R.H.S. }
Conclusion: Since the L.H.S. equals the R.H.S., the function y = Ax satisfies the given differential equation .
Thus, it is a solution.
Q6: Verify that the given function y = x \sin x is a solution of the differential equation xy' = y + x\sqrt{x^2 - y^2} (where x \ne 0 and x > y or x < -y ).
Solution:
Step 1: Write down the given function
Given function: y = x \sin x … (1)
Step 2: Find the First Derivative (y')
Differentiating equation (1) with respect to x
y' = \frac{d}{dx} (x) \cdot \sin x + x \cdot \frac{d}{dx} (\sin x)
y' = 1 \cdot \sin x + x \cos x
y' = \sin x + x \cos x … (2)
Step 3: Evaluate the Left Hand Side (L.H.S.)
The L.H.S. of the differential equation is xy' .
Substitute the value of y' from equation (2):
\text{L.H.S. } = x(\sin x + x \cos x)
\text{L.H.S. } = x \sin x + x^2 \cos x
Step 4: Evaluate the Right Hand Side (R.H.S.)
The R.H.S. is y + x\sqrt{x^2 - y^2} .
Substitute y = x \sin x into the expression:
\text {R.H.S. } = x \sin x + x\sqrt{x^2 - (x \sin x)^2}
\text{R.H.S. } = x \sin x + x\sqrt{x^2 - x^2 \sin^2 x}
\text{R.H.S. } = x \sin x + x\sqrt{x^2(1 - \sin^2 x)}
\text{R.H.S. } = x \sin x + x\sqrt{x^2 \cos^2 x)}
\text{R.H.S. } = x \sin x + x(x \cos x)
\text{R.H.S. } = x \sin x + x^2 \cos x
Step 5: Compare L.H.S. and R.H.S.
\text{L.H.S. } = \text{R.H.S. }
Conclusion: Since the L.H.S. equals the R.H.S., the function y = x \sin x satisfies the given differential equation .
Thus, it is a solution.
Q7: Verify that the given function xy = \log y + C is a solution of the differential equation y' = \frac{y^2}{1-xy} , (xy \ne 1) .
Solution:
Step 1: Write down the given function
Given function: xy = \log y + C … (1)
Step 2: Differentiate with respect to x
\frac{d}{dx}(xy) = \frac{d}{dx} (\log y + C)
x \cdot y' + y \cdot 1 = \frac{1}{y} \cdot y'
Step 3: Solve for y' Group the terms containing y' on one side:
y = \frac{1}{y}y' - xy'
y = y' (\frac{1}{y} - x)
y = y' (\frac{1 -xy}{y})
y' = \frac{y^2}{1 - xy}
Step 4: Compare with the Differential Equation
The derived equation y' = \frac{y^2}{1 - xy} matches the given differential equation exactly.
Conclusion: The function xy = \log y + C satisfies the given differential equation .
Thus, it is a solution.
Q8: Verify that the given function y - \cos y = x is a solution of the differential equation (y \sin y + \cos y + x)y' = y .
Solution:
Step 1: Write down the given function
Given function: y - \cos y = x … (1)
Step 2: Differentiate with respect to x
\frac{d}{dx} (y) - \frac{d}{dx} (\cos y) = \frac {d}{dx} (x)
y' - (-\sin y)y' = 1
y' + y' \sin y = 1
y' (1 + \sin y) = 1
y' = \frac{1}{1 + \sin y} … (2)
Step 3: Evaluate the Left Hand Side (L.H.S.)
The L.H.S. of the differential equation is (y \sin y + \cos y + x)y' .
From equation (1), substitute x = y - \cos y into the parenthesis:
\text{L.H.S. } = (y \sin y + \cos y + (y - \cos y))y'
\text{L.H.S. } = (y \sin y + y)y'
\text{L.H.S. } = y(\sin y + 1)y'
Now, substitute the value of y' from equation (2):
\text{L.H.S. } = y(\sin y + 1) \cdot \frac{1}{1 + \sin y }
\text{L.H.S. } = y
Step 4: Compare with Right Hand Side (R.H.S.)
The R.H.S. of the given equation is y .
\text {L.H.S. } = text {R.H.S. }
Conclusion: Since the L.H.S. equals the R.H.S., the function y - \cos y = x satisfies the given differential equation .
Thus, it is a solution.
Q9: Verify that the given function x + y = \tan^{-1} y is a solution of the differential equation y^2 y' + y^2 + 1 = 0 .
Solution:
Step 1: Write down the given function
Given function: x + y = \tan^{-1} y … (1)
Step 2: Differentiate with respect to x
\frac{d}{dx} (x + y) = \frac{d}{dx}(\tan^{-1} y)
1 + y' = \frac{1}{1 + y^2} \cdot y'
Step 3: Rearrange to match the Differential Equation form
Group y' terms on the right side:
1 = \frac{1}{1 + y^2}y' - y'
1 = y'(\frac{1}{1 + y^2} - 1)
1 = y'(\frac{1 - (1 + y^2)}{1 + y^2})
1 = y' (\frac{-y^2}{1 + y^2})
1 + y^2 = -y^2 \cdot y'
y^2 \cdot y' + y^2 + 1= 0
Step 4: Conclusion The derived equation matches the given differential equation.
Thus, the function is a solution.
Q10: Verify that the given function y = \sqrt{a^2 - x^2} is a solution of the differential equation x + y \frac{dy}{dx} = 0 , (y \ne 0) .
Solution:
Step 1: Write down the given function
Given function: y = \sqrt{a^2 - x^2} … (1)
Step 2: Find the First Derivative (\frac{dy}{dx})
Differentiating equation (1) with respect to x :
\frac{dy}{dx} = \frac{d}{dx}(a^2 - x^2)^{\frac{1}{2}}
\frac{dy}{dx} = \frac{1}{2}(a^2 - x^2)^{-\frac{1}{2}} \cdot \frac{d}{dx}(a^2 - x^2)
\frac{dy}{dx} = \frac{1}{2\sqrt{a^2 - x^2}} \cdot (-2x)
\frac{dy}{dx} = \frac{-x}{\sqrt{a^2 - x^2}}
Since y = \sqrt{a^2 - x^2} , we can substitute y in the denominator:
\frac{dy}{dx} = \frac{-x}{y} … (2)
Step 3: Substitute values into the Differential Equation
The L.H.S. is x + y \frac{dy}{dx} .
Substitute the value of \frac{dy}{dx} from equation (2):
\text{L.H.S. } = x + y (\frac{-x}{y})
\text{L.H.S. } = x - x
\text{L.H.S. } = 0
Step 4: Compare with Right Hand Side
The R.H.S. of the given equation is 0.
\text{L.H.S. } = \text{R.H.S }
Conclusion: Since the L.H.S. equals the R.H.S., the function y = \sqrt{a^2 - x^2} satisfies the given differential equation.
Thus, it is a solution.
Q11: The number of arbitrary constants in the general solution of a differential equation of fourth order are:
(A) 0
(B) 2
(C) 3
(D) 4
Solution:
The general solution of a differential equation is the solution that contains arbitrary constants (parameters).
The number of arbitrary constants in the general solution is exactly equal to the order of the differential equation.
The question mentions that the differential equation is of fourth order. Order = 4.
Since the order is 4, the number of arbitrary constants in its general solution must be 4.
Final Answer: The correct option is (D) .
Q12: The number of arbitrary constants in the particular solution of a differential equation of third order are:
(A) 3
(B) 2
(C) 1
(D) 0
Solution:
A particular solution is obtained from the general solution by giving specific values to the arbitrary constants.
These specific values are usually determined by initial or boundary conditions given in the problem.
Once specific values are assigned to the constants, they are no longer “arbitrary.” They become fixed numbers.
Therefore, a particular solution contains zero arbitrary constants, regardless of the order of the differential equation.
Even though the order is 3 (which would mean 3 arbitary constants in the general solution), the question asks about the particular solution.
Thus, the number of arbitrary constants is 0.
Final Answer: The correct option is (D) .