NCERT Solutions for Differential Equations Exercise 9.6 Solutions

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Class 12 Maths NCERT Solutions Chapter 9 Differential Equations Exercise 9.6 is provided in the article. Class 12 Chapter 9 Differential Equations Exercises are based on solving the linear differential equations.

Download PDF: NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.6

Read More: NCERT Solutions For Class 12 Mathematics Chapter 9 Differential Equations

Also check other Exercise Solutions of Class 12 Maths Chapter 9 Differential Equations

Also Read:

Chapter 9 Differential Equations Important Topics
Differential Equations Applications	Difference between parametric and nonparametric	Partial Differential Equation
Ordinary differential equations	Differential Equations Formula	Differential Calculus
Partial Derivative	How to Solve Linear Differential Equation	Homogeneous Differential Equation

Also Read:

CBSE Class 12 Mathematics Study Guides
Algebra	Arithmetic	Calculus
Mensuration	Trigonometry	Geometry
Permutation and Combination	Math Study Notes	Math MCQs
NCERT Solutions for Class 12 Maths	Math Formula	Difference between in Maths

CBSE CLASS XII Related Questions

1.
If \( xy = e^{x - y} \), then find \( \frac{dy}{dx} \).
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2.
Find the vector and cartesian equations of the line passing through the point of intersection of the lines \( \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j}) \) and \( \vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k}) \) and parallel to the line \( \frac{x - 1}{-2} = \frac{7 - y}{-3} = z \).
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3.
Find:

If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).
- \(p = -\frac{2}{3}, \, q = \frac{5}{3}\)
- \(p = -\frac{8}{3}, \, q = \frac{20}{3}\)
- \(p = \frac{20}{3}, \, q = -\frac{8}{3}\)
- \(p = 0, \, q = 0\)
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Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]
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A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.
6.
Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).
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