NCERT Solutions
Class 12 - Mathematics Part-2 - Chapter 9: DIFFERENTIAL EQUATIONS

Exercise 9.2

Verify that the given function is a solution of the corresponding differential equation:

Given: \(y = e^x + 1\)

Differential equation: \(y'' - y' = 0\)

Verify that the given function is a solution of the corresponding differential equation:

Given: \(y = x^2 + 2x + C\)

Differential equation: \(y' - 2x - 2 = 0\)

Verify that the given function is a solution of the corresponding differential equation:

Given: \(y = \cos x + C\)

Differential equation: \(y' + \sin x = 0\)

Verify that the given function is a solution of the corresponding differential equation:

Given: \(y = \sqrt{1 + x^2}\)

Differential equation: \(y' = \frac{xy}{1 + x^2}\)

Verify that the given function is a solution of the corresponding differential equation:

Given: \(y = Ax\)

Differential equation: \(xy' = y\) \((x \ne 0)\)

Verify that the given function is a solution of the corresponding differential equation:

Given: \(y = x\sin x\)

Differential equation: \(xy' = y + x\sqrt{x^2 - y^2}\) \((x \ne 0\ \text{and}\ x > y\ \text{or}\ x < -y)\)

Verify that the given function is a solution of the corresponding differential equation:

Given: \(xy = \log y + C\)

Differential equation: \(y' = \frac{y^2}{1 - xy}\) \((xy \ne 1)\)

Verify that the given function is a solution of the corresponding differential equation:

Given: \(y - \cos y = x\)

Differential equation: \((y\sin y + \cos y + x)\,y' = y\)

Verify that the given function is a solution of the corresponding differential equation:

Given: \(x + y = \tan^{-1}y\)

Differential equation: \(y^2y' + y^2 + 1 = 0\)

Verify that the given function is a solution of the corresponding differential equation:

Given: \(y = \sqrt{a^2 - x^2},\ x \in (-a, a)\)

Differential equation: \(x + y\frac{dy}{dx} = 0\) \((y \ne 0)\)