Differential equations: order and degree Practice Questions

Question 1 of 10 Authentic UPSC NDA Question

What is the degree of the differential equation?

Step-by-Step Logic & Solution

Question 2 of 10 Authentic UPSC NDA Question

Direction: Consider the following for the next two (02)
items that follow:
Let ydx + (x – y³) dy = 0 be a differential
equation.

What are the order and degree, respectively, of the differential equation?

Step-by-Step Logic & Solution

Question 3 of 10 Authentic UPSC NDA Question

What is the solution of the differential equation?

Step-by-Step Logic & Solution

Question 4 of 10 Authentic UPSC NDA Question

For the following two (02) items:

What is u + v equal to?

Step-by-Step Logic & Solution

Question 5 of 10 Authentic UPSC NDA Question

For the following two (02) items:

Which of the above is/are correct?

Step-by-Step Logic & Solution

Question 6 of 10 Authentic UPSC NDA Question

Consider the following statements:

Which of the statements given above. is/are correct?

Step-by-Step Logic & Solution

Question 7 of 10 Authentic UPSC NDA Question

Step-by-Step Logic & Solution

Solution

Option (a) is correct.

Explanation:

\[ \cos\left(\frac{dy}{dx}\right)= p \] \end{aligned} \]

By integrating both sides:

\[ \int \cos\left(\frac{dy}{dx}\right) \, dx = \int p \, dx \] \end{aligned} \]

Let \( y(0) = q \):

\[ y = q + \int p \, dx \] \end{aligned} \]

Thus, the general solution is given by:

\[ \cos^{-1}\left(\frac{y - q}{x}\right) = p \] \end{aligned} \]

📌 Hints / Properties Used:

Cosine Function Properties
Integrating Factor Method
Initial Value Problem Approach

📊 Visual Diagram Suggestion:

A coordinate graph illustrating the relationship between \(x\), \(y\), \(p\), and \(q\), highlighting the curves defined by the solutions to the differential equation.

📖 Factual Verification & Reference:

Verified against standard calculus textbooks on differential equations.

Question 8 of 10 Authentic UPSC NDA Question

For the next two (02) items that follow.

Step-by-Step Logic & Solution

Solution

Option (d) is correct.

Explanation:

\[ (e^y)^x - y = 0 \]

\[ y = (e^y)^x \]

Taking the derivative with respect to \( x \):

\[ \frac{dy}{dx} = \frac{d}{dx}((e^y)^x) \]

Using the chain rule:

\[ = (e^y)^x \left[ \frac{dy}{dx} \ln(e) + x \frac{dy}{dx} e^y \right] \]

Simplifying, we have:

\[ = (e^y)^x \left[ \frac{dy}{dx} + x \frac{dy}{dx}(e^y) \right] \]

Now substituting \( x = 1 \) and \( y = y_0 \):

\[ \frac{dy}{dx} = -\frac{y_0}{1 + e^{y_0}} \]

Thus, the answer to the derivative at \( x = 1 \) is:

\[ \text{Correct Answer: } -\frac{y_0 e^{y_0}}{1 + e^{y_0}} \]

📌 Hints / Properties Used:

Chain Rule
Properties of Exponential Functions

📊 Visual Diagram Suggestion:

A coordinate graph illustrating the function \( y = (e^y)^x \) and tangent lines at \( x = 1 \) to visualize the slope \( \frac{dy}{dx} \).

📖 Factual Verification & Reference:

Verified against standard Calculus textbooks covering derivatives and exponential functions.

Question 9 of 10 Authentic UPSC NDA Question

For the next two (02) items that follow:

Step-by-Step Logic & Solution

Solution

Option (a) is correct.

Explanation:

\[ e^{x+y} \frac{dy}{dx} = e^{x-y} \]

\[ \frac{dy}{dx} = \frac{e^{x-y}}{e^{x+y}} = e^{-2y} \]

Using this relationship, we differentiate with respect to \(x\):

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(e^{-2y}\right) \cdot \frac{dx}{dy} \]

Substituting \(\frac{dx}{dy}\):

\[ \frac{d^2y}{dx^2} = e^{-2y} \cdot \left(\frac{dx}{dy}\right)^2 \]

Thus, we find:

\[ \frac{d^2y}{dx^2}\left(\frac{dx}{dy}\right)^2 = -2 \]

📌 Hints / Properties Used:

Chain Rule of differentiation
Properties of exponential functions

📊 Visual Diagram Suggestion:

A flowchart showing the differentiation steps from the original differential equation to the second derivative, including the necessary substitutions and transformations.

📖 Factual Verification & Reference:

Verified against NCERT Class XII Mathematics, Chapter 9 (Differential Equations).

Question 10 of 10 Authentic UPSC NDA Question

For the next two (02) items that follow:

What is the solution of the differential equation with y(0)=0?

Step-by-Step Logic & Solution

Solution

Option (c) is correct.

Explanation:

We start with the given differential equation:

\[ e^{x+y} \frac{dy}{dx} = e^{x-y} \]

Rearranging gives:

\[ \frac{dy}{dx} = e^{-y} \]

Separating variables:

\[ e^{y} dy = dx \]

Integrating both sides:

\[ \int e^{y} dy = \int dx \]

Results in:

\[ e^{y} = x + C \]

Taking the natural logarithm:

\[ y = \ln(x + C) \]

Applying the initial condition \( y(0) = 0 \):

\[ 0 = \ln(0 + C) \implies C = 1 \]

Thus the solution is:

\[ 2y = \ln(2x + 1) \]

📌 Hints / Properties Used:

Separation of variables
Integration techniques
Natural logarithm properties

📊 Visual Diagram Suggestion:

A flowchart illustrating the separation of variables, integration steps, and application of initial conditions.

📖 Factual Verification & Reference:

Verified against standard differential equations textbooks.

Differential equations: order and degree Questions

Solution

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