Class CBSE Class 12 Mathematics Differential Equations Q #917
KNOWLEDGE BASED
APPLY
3 Marks 2023 SA
Find the general solution of the differential equation \(e^{x}\tan y~dx+(1-e^{x})\sec^{2}y~dy=0\).
Prev Next

AI Tutor Explanation

Powered by Gemini

Step-by-Step Solution

  1. Given differential equation: \(e^{x}\tan y~dx+(1-e^{x})\sec^{2}y~dy=0\)

  2. Rearrange the equation to separate variables:

    \(e^{x}\tan y~dx = -(1-e^{x})\sec^{2}y~dy\)

    \(\frac{e^{x}}{1-e^{x}}dx = -\frac{\sec^{2}y}{\tan y}dy\)

  3. Integrate both sides:

    \(\int \frac{e^{x}}{1-e^{x}}dx = -\int \frac{\sec^{2}y}{\tan y}dy\)

  4. For the left side, let \(u = 1-e^{x}\), then \(du = -e^{x}dx\). So, \(\int \frac{e^{x}}{1-e^{x}}dx = -\int \frac{1}{u}du = -\ln|u| = -\ln|1-e^{x}|\)

  5. For the right side, let \(v = \tan y\), then \(dv = \sec^{2}y~dy\). So, \(-\int \frac{\sec^{2}y}{\tan y}dy = -\int \frac{1}{v}dv = -\ln|v| = -\ln|\tan y|\)

  6. Combine the results:

    -\(\ln|1-e^{x}| = -\ln|\tan y| + C\)

    \(\ln|1-e^{x}| = \ln|\tan y| - C\)

    \(\ln|1-e^{x}| - \ln|\tan y| = -C\)

    \(\ln\left|\frac{1-e^{x}}{\tan y}\right| = -C\)

  7. Exponentiate both sides:

    \(\frac{1-e^{x}}{\tan y} = e^{-C} = K\) (where K is a constant)

  8. Therefore, the general solution is:

    \(1-e^{x} = K\tan y\)

Correct Answer: \(1-e^{x} = K\tan y\)

AI generated content. Review strictly for academic accuracy.

Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply their knowledge of differential equations and integration techniques to solve the given problem.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to follow a specific procedure to solve the differential equation, including separation of variables and integration.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's understanding and application of standard methods for solving differential equations, a core topic in the syllabus.

More from this Chapter

MCQ_SINGLE
The order of the differential equation \(\frac{d^{4}y}{dx^{4}}-sin(\frac{d^{2}y}{dx^{2}})=5\) is:
SA
Find the particular solution of the differential equation:$$\frac{dy}{dx} + \sec^{2}x \cdot y = \tan x \cdot \sec^{2}x$$given that $y(0) = 0$.
MCQ_SINGLE
Which of the following is not a homogeneous function of \(x\) and \(y\) ?
SA
Solve the differential equation given by:$$x \, dy - y \, dx - \sqrt{x^{2} + y^{2}} \, dx = 0$$
SUBJECTIVE
(i) Write the order and degree of the given differential equation. (ii) Substituting $V = \pi r^3$ and $S = 2\pi r^2$, we get the differential equation $\frac{dr}{dt} = \frac{2}{3}k$. Solve it, given that $r(0) = 5$ mm. (iii) (a) If it is given that $r = 3$ mm when $t = 1$ hour, find the value of k. Hence, find t for $r = 0$ mm. OR (iii) (b) If it is given that $r = 1$ mm when $t = 1$ hour, find the value of k. Hence, find t for $r = 0$ mm.
View All Questions