Question:
The differential equation satisfied by the family of curve \( y = ax \cos \left( \frac{1}{x} + b \right) \), where \( a, b \) are parameters, is:
The differential equation satisfied by the family of curve \( y = ax \cos \left( \frac{1}{x} + b \right) \), where \( a, b \) are parameters, is:
Show Hint
When differentiating a product of functions like this, carefully apply the product rule and simplify the resulting expression step by step.
Updated On: Apr 22, 2026
- \( x^2 y_2 + y = 0 \)
- \( x^4 y_2 + y = 0 \)
- \( x y_2 - y = 0 \)
- \( x^4 y_2 - y = 0 \)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Start with the given equation.
We are given the family of curves defined by: \[ y = ax \cos \left( \frac{1}{x} + b \right) \] where \( a \) and \( b \) are constants. To eliminate the parameters \( a \) and \( b \), we differentiate this equation with respect to \( x \).
Step 2: First differentiation to find \( y_1 \).
Differentiating both sides with respect to \( x \), we apply the product rule: \[ \frac{dy}{dx} = a \cos \left( \frac{1}{x} + b \right) - a x \sin \left( \frac{1}{x} + b \right) \cdot \left( -\frac{1}{x^2} \right) \] Simplifying: \[ y_1 = a \cos \left( \frac{1}{x} + b \right) + a \frac{\sin \left( \frac{1}{x} + b \right)}{x} \]
Step 3: Second differentiation to find \( y_2 \).
Now, differentiate \( y_1 \) again with respect to \( x \) to find \( y_2 \). Using the product rule again on both terms: \[ y_2 = -a \frac{\sin \left( \frac{1}{x} + b \right)}{x} + a \frac{\cos \left( \frac{1}{x} + b \right)}{x^2} - a \left( \frac{\cos \left( \frac{1}{x} + b \right)}{x^2} + 2 \frac{\sin \left( \frac{1}{x} + b \right)}{x^3} \right) \] Simplifying: \[ y_2 = -a \frac{\sin \left( \frac{1}{x} + b \right)}{x} + a \frac{\cos \left( \frac{1}{x} + b \right)}{x^2} - a \frac{\cos \left( \frac{1}{x} + b \right)}{x^2} - 2a \frac{\sin \left( \frac{1}{x} + b \right)}{x^3} \] We can see that the terms involving \( \cos \) cancel out.
Step 4: Substituting back to form the differential equation.
Now, we return to the original equation \( y = ax \cos \left( \frac{1}{x} + b \right) \), which gives: \[ y = ax \cos \left( \frac{1}{x} + b \right) \] and \[ y_2 = -2a \frac{\sin \left( \frac{1}{x} + b \right)}{x^3} \] From the above relationships, we can see that: \[ x^4 y_2 + y = 0 \] Thus, the differential equation satisfied by the family of curves is: \[ \boxed{x^4 y_2 + y = 0} \]
We are given the family of curves defined by: \[ y = ax \cos \left( \frac{1}{x} + b \right) \] where \( a \) and \( b \) are constants. To eliminate the parameters \( a \) and \( b \), we differentiate this equation with respect to \( x \).
Step 2: First differentiation to find \( y_1 \).
Differentiating both sides with respect to \( x \), we apply the product rule: \[ \frac{dy}{dx} = a \cos \left( \frac{1}{x} + b \right) - a x \sin \left( \frac{1}{x} + b \right) \cdot \left( -\frac{1}{x^2} \right) \] Simplifying: \[ y_1 = a \cos \left( \frac{1}{x} + b \right) + a \frac{\sin \left( \frac{1}{x} + b \right)}{x} \]
Step 3: Second differentiation to find \( y_2 \).
Now, differentiate \( y_1 \) again with respect to \( x \) to find \( y_2 \). Using the product rule again on both terms: \[ y_2 = -a \frac{\sin \left( \frac{1}{x} + b \right)}{x} + a \frac{\cos \left( \frac{1}{x} + b \right)}{x^2} - a \left( \frac{\cos \left( \frac{1}{x} + b \right)}{x^2} + 2 \frac{\sin \left( \frac{1}{x} + b \right)}{x^3} \right) \] Simplifying: \[ y_2 = -a \frac{\sin \left( \frac{1}{x} + b \right)}{x} + a \frac{\cos \left( \frac{1}{x} + b \right)}{x^2} - a \frac{\cos \left( \frac{1}{x} + b \right)}{x^2} - 2a \frac{\sin \left( \frac{1}{x} + b \right)}{x^3} \] We can see that the terms involving \( \cos \) cancel out.
Step 4: Substituting back to form the differential equation.
Now, we return to the original equation \( y = ax \cos \left( \frac{1}{x} + b \right) \), which gives: \[ y = ax \cos \left( \frac{1}{x} + b \right) \] and \[ y_2 = -2a \frac{\sin \left( \frac{1}{x} + b \right)}{x^3} \] From the above relationships, we can see that: \[ x^4 y_2 + y = 0 \] Thus, the differential equation satisfied by the family of curves is: \[ \boxed{x^4 y_2 + y = 0} \]
Was this answer helpful?
0
0
Top MET Mathematics Questions
- $\int\limits_{0}^{8}| x -5| d x$ is equal to
- The equation $3^{3x + 4} = 9^{2x - 2},\ x>0$ has the solution
- A parallelogram is constructed on the vectors $\mathbf{a} = 3\alpha - \beta$, $\mathbf{b} = \alpha + 3\beta$. If $|\alpha| = |\beta| = 2$ and the angle between $\alpha$ and $\beta$ is $\dfrac{\pi}{3}$, then the length of a diagonal of the parallelogram is
- Every group of order 7 is
- Let $U_{n} = 2 + 2^{3} + 2^{5} + \cdots + 2^{2n+1}$ and $V_{n} = 1 + 4 + 4^{2} + \cdots + 4^{n-1}$. Then $\displaystyle\lim_{n \to \infty} \dfrac{U_n}{V_n}$ is equal to
View More Questions
Top MET Differential equations Questions
- The general solution of $x\sqrt{1+y^{2}}\,dx + y\sqrt{1+x^{2}}\,dy = 0$ is
- By eliminating the arbitrary constants $A$ and $B$ from $y = Ax^{2} + Bx$, the differential equation obtained is
- If $x(1+y^{2})\,dx + y(1+x^{2})\,dy = 0$ and $y(0) = 1$, then $x^{2}y^{2} + x^{2} + y^{2}$ equals
- The differential equation of the family of curves $y = a\cos\mu x + b\sin\mu x$, where $a$ and $b$ are arbitrary constants, is
- The equation of the curve passing through the origin and satisfying $\dfrac{dy}{dx} = (x - y)^{2}$ is
View More Questions
Top MET Questions
- A coil of 100 turns and area $2 \times 10^{-2}m^2 $ is pivoted about a vertical diameter in a uniform magnetic field and carries a current of 5A. When the coil is held with its plane in north-south direction, it experiences a couple of 0.33 Nm. When the plane is east-west, the corresponding couple is 0.4 Nm, the value of magnetic induction is [Neglect earth's magnetic field]
- $\int\limits_{0}^{8}| x -5| d x$ is equal to
- Radiation, with wavelength 6561 $?$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{-4} T$. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to :
- In intrinsic semiconductor at room temperature number of electrons and holes are
- Which of the following product is formed in the reaction $ CH_3MgBr {->[(i)CO_2][(ii)H_2O]} ? $
View More Questions