NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.2


Chapter 9 Ex.9.2 Question 1

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

\(y = {e^x} + 1{\rm{ }}:{\rm{ }}y'' - y' = 0\)

 

Solution

 

\(y = {e^x} + 1\)

\[\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {{e^x} + 1} \right)\\ &\Rightarrow \; y' = {e^x}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\\frac{d}{{dx}}\left( {y'} \right) &= \frac{d}{{dx}}\left( {{e^x}} \right)\\ &\Rightarrow \; y'' = {e^x}\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}\]

From \(\left( 1 \right)\) and \(\left( 2 \right)\)

\[\begin{align}y'' - y' &= {e^x} - {e^x}\\ &= 0\end{align}\]

Thus, the given function is the solution of corresponding differential equation.

Chapter 9 Ex.9.2 Question 2

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

\(y = {x^2} + 2x + c{\rm{ }}:{\rm{ }}y' - 2x - 2 = 0\)

 

Solution

 

\[\begin{align}y &= {x^2} + 2x + c\\y' &= \frac{d}{{dx}}\left( {{x^2} + 2x + c} \right)\\ \Rightarrow \; y' &= 2x + 2\end{align}\]

Therefore,

\[\begin{align}y' - 2x - 2& = 2x + 2 - 2x - 2\\& = 0\end{align}\]

Thus, the given function is the solution of the differential equation.

Chapter 9 Ex.9.2 Question 3

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

\(y = \cos x + C \quad : \quad y' + \sin x = 0\)

 

Solution

 

\[\begin{align}y &= \cos x + C\\y' &= \frac{d}{{dx}}\left( {\cos x + C} \right)\\ \Rightarrow \; y' &= - \sin x\end{align}\]

Therefore,

\[\begin{align}y' + \sin x &= - \sin x + \sin x\\ &= 0\end{align}\]

Thus, the given function is the solution of the differential equation.

Chapter 9 Ex.9.2 Question 4

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

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

 

Solution

 

\[\begin{align}y &= \sqrt {1 + {x^2}} \\y' &= \frac{d}{{dx}}\left( {\sqrt {1 + {x^2}} } \right)\\& = \frac{1}{{2\sqrt {1 + {x^2}} }}.\frac{d}{{dx}}\left( {1 + {x^2}} \right)\\ &= \frac{{2x}}{{2\sqrt {1 + {x^2}} }}\\& = \frac{x}{{2\sqrt {1 + {x^2}} }}\\ &= \frac{x}{{\left( {1 + {x^2}} \right)}} \times \sqrt {1 + {x^2}} \\ &= \frac{x}{{\left( {1 + {x^2}} \right)}}.y\\ &= \frac{{xy}}{{1 + {x^2}}}\end{align}\]

Thus, the given function is the solution of the differential equation.

Chapter 9 Ex.9.2 Question 5

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

\(y = Ax \quad:\quad xy' = y\left( {x \ne 0} \right)\)

 

Solution

 

\[\begin{align}y &= Ax\\y' &= \frac{d}{{dx}}\left( {Ax} \right)\\ &= A\end{align}\]

Therefore,

\[\begin{align}xy' &= xA\\& = Ax\\ &= y\end{align}\]

Thus, the given function is the solution of the differential equation.

Chapter 9 Ex.9.2 Question 6

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

\(y = x\sin x \quad : \quad xy' = y + x\sqrt {{x^2} - {y^2}} \left( {x \ne 0 \text{ and }x > y} \right)\;or\;x < - y{\rm{ }}\)

 

Solution

 

\[\begin{align}y &= x\sin x\\y' &= \frac{d}{{dx}}\left( {x\sin x} \right)\\ &= \sin x.\frac{d}{{dx}}\left( x \right) + x.\frac{d}{{dx}}\left( {\sin x} \right)\\ &= \sin x + x\cos x\end{align}\]

Therefore,

\[\begin{align}xy' &= x\left( {\sin x + x\cos x} \right)\\ &= x\sin x + {x^2}\cos x\\ &= y + {x^2}.\sqrt {1 - {{\sin }^2}x} \\ &= y + {x^2}\sqrt {1 - {{\left( {\frac{y}{x}} \right)}^2}} \\ &= y + x\sqrt {{x^2} - {y^2}}\end{align}\]

Thus, the given function is the solution of the differential equation.

Chapter 9 Ex.9.2 Question 7

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

\(xy = \log y + C \quad : \quad y' = \frac{{{y^2}}}{{1 - xy}}\left( {xy \ne 1} \right)\)

 

Solution

 

\[\begin{align}&xy = \log y + C\\ &\Rightarrow \; \frac{d}{{dx}}\left( {xy} \right) = \frac{d}{{dx}}\left( {\log y} \right)\\ &\Rightarrow \; y\frac{d}{{dx}}\left( x \right) + x.\frac{{dy}}{{dx}} = \frac{1}{y}\frac{{dy}}{{dx}}\\ &\Rightarrow \; y + xy' = \frac{1}{y}.y'\\ &\Rightarrow \; {y^2} + xyy' = y'\\ &\Rightarrow \; \left( {xy - 1} \right)y' = - {y^2}\\ &\Rightarrow \; y' = \frac{{{y^2}}}{{1 - xy}}\end{align}\]

Thus, the given function is the solution of the differential equation.

Chapter 9 Ex.9.2 Question 8

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

\(y - \cos y = x \quad : \quad \left( {y\sin y + \cos y + x} \right)y' = y\)

 

Solution

 

\[\begin{align}&y - \cos y = x\\& \Rightarrow \; \frac{{dy}}{{dx}} - \frac{d}{{dx}}\left( {\cos y} \right) = \frac{d}{{dx}}(x)\\& \Rightarrow \; y' - \left( { - \sin y} \right).y' = 1\\ &\Rightarrow \; y'\left( {1 + \sin y} \right) = 1\\& \Rightarrow \; y' = \frac{1}{{1 + \sin y}}\end{align}\]

Therefore,

\[\begin{align}\left( {y\sin y + \cos y + x} \right)y' &= \left( {y\sin y + \cos y + y - \cos y} \right) \times \frac{1}{{1 + \sin y}}\\ &= y\left( {1 + \sin y} \right).\frac{1}{{1 + \sin y}}\\& = y\end{align}\]

Thus, the given function is the solution of the differential equation.

Chapter 9 Ex.9.2 Question 9

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

\(x + y = {\tan ^{ - 1}}y \quad : \quad {y^2}y' + {y^2} + 1 = 0\)

 

Solution

 

\[\begin{align}&x + y = {\tan ^{ - 1}}y\\& \Rightarrow \; \frac{d}{{dx}}\left( {x + y} \right) = \frac{d}{{dx}}\left( {{{\tan }^{ - 1}}y} \right)\\& \Rightarrow \; 1 + y' = \left[ {\frac{1}{{1 + {y^2}}}} \right]y'\\ &\Rightarrow \; y'\left[ {\frac{1}{{1 + {y^2}}} - 1} \right] = 1\\& \Rightarrow \; y'\left[ {\frac{{1 - \left( {1 + {y^2}} \right)}}{{1 + {y^2}}}} \right] = 1\\& \Rightarrow \; y'\left[ {\frac{{ - {y^2}}}{{1 + {y^2}}}} \right] = 1\\ &\Rightarrow \; y' = \frac{{ - \left( {1 + {y^2}} \right)}}{{{y^2}}}\end{align}\]

Therefore,

\[\begin{align}{y^2}y' + {y^2} + 1& = {y^2}\left[ {\frac{{ - (1 + {y^2})}}{{{y^2}}}} \right] + {y^2} + 1\\ &= - 1 - {y^2} + {y^2} + 1\\ &= 0\end{align}\]

Thus, the given function is the solution of the differential equation.

Chapter 9 Ex.9.2 Question 10

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

\(y = \sqrt {{a^2} - {x^2}} \;x \in \left( { - a,a} \right) \quad : \quad x + y\frac{{dy}}{{dx}} = 0\left( {y \ne 0} \right)\)

 

Solution

 

\[\begin{align}&y = \sqrt {{a^2} - {x^2}} \\&\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {\sqrt {{a^2} - {x^2}} } \right)\\& \Rightarrow \; \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {{a^2} - {x^2}} }}.\frac{d}{{dx}}\left( {{a^2} - {x^2}} \right)\\& \Rightarrow \; \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {{a^2} - {x^2}} }}.\left( { - 2x} \right)\\ &\Rightarrow \; \frac{{dy}}{{dx}} = \frac{{ - x}}{{\sqrt {{a^2} - {x^2}} }}\end{align}\]

Therefore,

\[\begin{align}x + y\frac{{dy}}{{dx}} &= x + \sqrt {{a^2} - {x^2}} \times \frac{{ - x}}{{\sqrt {{a^2} - {x^2}} }}\\ &= x - x\\& = 0\end{align}\]

Thus, the given function is the solution of the differential equation.

Chapter 9 Ex.9.2 Question 11

The numbers of arbitrary constants in the general solution of a differential equation of fourth order are:

(A) \(0\)

(B) \(2\)

(C) \(3\)

(D) \(4\)

 

Solution

 

We know that the number of constants in the general solution of a differential equation of order \(n\) is equal to its order.

The number of constants in general equation of fourth order differential equation is \(4.\)

Thus, the correct option is D.

Chapter 9 Ex.9.2 Question 12

The numbers of arbitrary constants in the particular solution of a differential equation of third order are:

(A) \(3\)

(B) \(2\)

(C) \(1\)

(D) \(0\)

 

Solution

 

In a particular solution of a differential equation, there are no arbitrary constants.

Thus, the correct option is D.

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