NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.3


Chapter 9 Ex.9.3 Question 1

Form a differential equation representing the given family of curves by eliminating arbitrary constants \(a\) and \(b\).

\(\frac{x}{a} + \frac{y}{b} = 1\)

 

Solution

 

\[\begin{align}&\frac{x}{a} + \frac{y}{b} = 1\\ &\Rightarrow \; \frac{1}{a} + \frac{1}{b}\frac{{dy}}{{dx}} = 0\\& \Rightarrow \; \frac{1}{a} + \frac{1}{b}y' = 0\\ &\Rightarrow \; 0 + \frac{1}{b}y'' = 0\\& \Rightarrow \; \frac{1}{b}y'' = 0\\ &\Rightarrow \; y'' = 0\end{align}\]

Thus, the differential equation of the given curve is \(y'' = 0\).

Chapter 9 Ex.9.3 Question 2

Form a differential equation representing the given family of curves by eliminating arbitrary constants \(a\) and \(b\).

\({y^2} = a\left( {{b^2} - {x^2}} \right)\)

 

Solution

 

\[\begin{align}&{y^2} = a\left( {{b^2} - {x^2}} \right)\\& \Rightarrow \; 2y\frac{{dy}}{{dx}} = a\left( { - 2x} \right)\\ &\Rightarrow \; 2yy' = - 2ax\\ &\Rightarrow \; yy' = - ax\\ &\Rightarrow \; y'.y' + yy'' = - a\\ &\Rightarrow \; \frac{{{{\left( {y'} \right)}^2} + yy''}}{{yy'}} = \frac{{ - a}}{{ - ax}}\\ &\Rightarrow \; xyy'' + x{\left( {y'} \right)^2} - yy' = 0\end{align}\]

Thus, the differential equation of the given curve is \(xyy'' + x{\left( {y'} \right)^2} - yy' = 0\).

Chapter 9 Ex.9.3 Question 3

Form a differential equation representing the given family of curves by eliminating arbitrary constants \(a\) and \(b\).

\(y = a{e^{3x}} + b{e^{ - 2x}}\)

 

Solution

 

\[\begin{align}&y = a{e^{3x}} + b{e^{ - 2x}}\\&y' = 3a{e^{3x}} - 2b{e^{ - 2x}} \qquad \ldots \left( 1 \right)\\&y'' = 9a{e^{3x}} + 4b{e^{ - 2x}} \qquad \ldots \left( 2 \right)\end{align}\]

Subtracting \(\left( 1 \right)\) from \(\left( 2 \right)\)

\[\begin{align}&y'' - y' = 9a{e^{3x}} + 4b{e^{ - 2x}} - 3a{e^{3x}} + 2b{e^{ - 2x}} \hfill \\&y'' - y' = 6a{e^{3x}} + 6b{e^{ - 2x}} \hfill \\&y'' - y' = 6\left( {a{e^{3x}} + b{e^{ - 2x}}} \right) \hfill \\&y'' - y' = 6y \qquad \left( {\because y = a{e^{3x}} + b{e^{ - 2x}}} \right) \hfill \\&y'' - y' - 6y = 0 \hfill \\\end{align} \]

Thus, the differential equation of the given curve is \(y'' - y' - 6y = 0\).

Chapter 9 Ex.9.3 Question 4

Form a differential equation representing the given family of curves by eliminating arbitrary constants \(a\) and \(b\).

\(y = {e^{2x}}\left( {a + bx} \right)\)

 

Solution

 

\[\begin{align}&y = {e^{2x}}\left( {a + bx} \right)\\&y' = 2{e^{2x}}\left( {a + bx} \right) + {e^{2x}}b\\ &\Rightarrow \; y' = {e^{2x}}\left( {2a + 2bx + b} \right)\\&y' - 2y = {e^{2x}}\left( {2a + 2bx + b} \right) - {e^{2x}}\left( {2a + 2bx} \right)\\& \Rightarrow \; y' - 2y = b{e^{2x}}  \qquad  \qquad  \ldots \left( 1 \right)\\ &\Rightarrow \; y'' - 2y' = 2b{e^{2x}}  \qquad  \quad  \ldots \left( 2 \right)\end{align}\]

Dividing \(\left( 2 \right)\) by \(\left( 1 \right)\)

\[\begin{align}&\frac{{y'' - 2y'}}{{y' - 2y}} = 2\\ &\Rightarrow \; y'' - 2y' = 2y' - 4y\\ &\Rightarrow \; y'' - 4y' + 4y = 0\end{align}\]

Thus, the differential equation of the given curve is \(y'' - 4y' + 4y = 0\).

Chapter 9 Ex.9.3 Question 5

Form a differential equation representing the given family of curves by eliminating arbitrary constants \(a\) and \(b\).

\(y = {e^{3x}}\left( {a\cos x + b\sin x} \right)\)

 

Solution

 

\[\begin{align}&y = {e^x}\left( {a\cos x + b\sin x} \right)\\&y' = {e^x}\left( {a\cos x + b\sin x} \right) + {e^x}\left( { - a\sin x + b\cos x} \right)\\ &\Rightarrow \; y' = {e^x}\left[ {\left( {a + b} \right)\cos x - \left( {a - b} \right)\sin x} \right]\\ &\Rightarrow \; y'' = {e^x}\left[ {\left( {a + b} \right)\cos x - \left( {a - b} \right)\sin x} \right] \\&\qquad+ {e^x}\left[ { - \left( {a + b} \right)\sin x - \left( {a - b} \right)\cos x} \right]\\&y'' = {e^x}\left[ {2b\cos x - 2a\sin x} \right]\\&y'' = 2{e^x}\left( {b\cos x - a\sin x} \right)\\ &\Rightarrow \; \frac{{y''}}{2} = {e^x}\left( {b\cos x - a\sin x} \right)\\&\frac{{y''}}{2} + y = {e^x}\left( {b\cos x - a\sin x} \right) + {e^x}\left( {a\cos x + b\sin x} \right)\\&y + \frac{{y''}}{2} = {e^x}\left[ {\left( {a + b} \right)\cos x - \left( {a - b} \right)\sin x} \right]\\ &\Rightarrow \; y + \frac{{y''}}{2} = y'\\& \Rightarrow \; 2y + y'' = 2y'\\& \Rightarrow \; y'' - 2y' + 2y = 0\end{align}\]

Thus, the differential equation of the given curve is \(y'' - 2y' + 2y = 0\).

Chapter 9 Ex.9.3 Question 6

Form the differential equation of the family of circles touching the \(y\)-axis at origin.

 

Solution

 

Centre of circle touching the \(y\)-axis at origin lies on the \(x\)-axis.

Let \(\left( {a,0} \right)\) be the centre of the circle.

Since it touches the \(y\)-axis at origin, its radius is \(a\).

Equation of the circle with centre \(\left( {a,0} \right)\) and radius \(a\) is

\[\begin{align}&{\left( {x - a} \right)^2} + {y^2} = {a^2}\\ &\Rightarrow \; {x^2} + {y^2} = 2ax \qquad \ldots \left( 1 \right)\end{align}\]

Differentiating equation \(\left( 1 \right)\) with respect to \(x\), we get:

\[\begin{align}&2x + 2yy' = 2a\\& \Rightarrow \; x + yy' = a\end{align}\]

Putting the value of \(a\) in equation \(\left( 1 \right)\)

\[\begin{align} &\Rightarrow \; {x^2} + {y^2} = 2\left( {x + yy'} \right)x\\& \Rightarrow \; 2xyy' + {x^2} = {y^2}\end{align}\]

Chapter 9 Ex.9.3 Question 7

Form the differential equation of the family of parabolas having vertex at origin and axis along positive \(y\)-axis.

 

Solution

 

The equation of the parabola having the vertex at origin and the axis along the positive \(y\)-axis is

\({x^2} = 4ay \qquad  \ldots \left( 1 \right)\).

Differentiating equation \(\left( 1 \right)\) with respect to \(x\), we get:

\[\begin{align}2x &= 4ay'\\ &\Rightarrow \; \frac{{2x}}{{{x^2}}} = \frac{{4ay'}}{{4ay}}\\& \Rightarrow \; \frac{2}{x} = \frac{{y'}}{y}\\& \Rightarrow \; xy' = 2y\\ &\Rightarrow \; xy' - 2y = 0\end{align}\]

Chapter 9 Ex.9.3 Question 8

Form the differential equation of the family of ellipses having foci on-axis and centre at origin.

 

Solution

 

The equation of the family of ellipses having foci on the \(y\)-axis and the centre at origin is as follows:

\(\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1 \qquad  \ldots \left( 1 \right)\)

Differentiating equation \((1)\) with respect to \(x\), we get:

\[\begin{align}&\frac{{2x}}{{{b^2}}} + \frac{{2yy'}}{{{a^2}}} = 0\\ &\Rightarrow \; \frac{x}{{{b^2}}} + \frac{{yy'}}{{{a^2}}} = 0  \qquad \dots \left( 2 \right)\end{align}\]

Differentiating equation \(\left( 2 \right)\) with respect to \(x\), we get:

\[\begin{align}&\frac{1}{{{b^2}}} + \frac{{y'.y' + y.y''}}{{{a^2}}} = 0\\ &\Rightarrow \; \frac{1}{{{b^2}}} + \frac{1}{{{a^2}}}\left( {{{y'}^2} + yy''} \right) = 0\\ &\Rightarrow \; \frac{1}{{{b^2}}} = - \frac{1}{{{a^2}}}\left( {{{y'}^2} + yy''} \right)\end{align}\]

Substituting this value in equation \(\left( 2 \right)\), we get:

\[\begin{align} &\Rightarrow \; x\left[ { - \frac{1}{{{a^2}}}\left( {{{y'}^2} + yy''} \right)} \right] + \frac{{yy'}}{{{a^2}}} = 0\\& \Rightarrow \; - x{{y'}^2} - xyy'' + yy' = 0\\ &\Rightarrow \; xyy'' + x{{y'}^2} - yy' = 0\end{align}\]

Chapter 9 Ex.9.3 Question 9

Form the differential equation of the family of hyperbolas having foci on \(x\)-axis and centre at origin.

 

Solution

 

The equation of the family of hyperbolas with the centre at origin and foci along the \(x\)-axis is

\(\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1  \qquad \ldots \left( 1 \right)\)

Differentiating equation \(\left( 1 \right)\) with respect to \(x\), we get:

\[\begin{align}&\frac{{2x}}{{{b^2}}} + \frac{{2yy'}}{{{a^2}}} = 0\\ &\Rightarrow \; \frac{x}{{{b^2}}} + \frac{{yy'}}{{{a^2}}} = 0 \qquad \ldots \left( 2 \right)\end{align}\]

Differentiating equation \(\left( 2 \right)\) with respect to \(x\), we get:

\[\begin{align}&\frac{1}{{{b^2}}} + \frac{{y'.y' + y.y''}}{{{a^2}}} = 0\\&\frac{1}{{{b^2}}} = - \frac{{y'.y' + y.y''}}{{{a^2}}}\end{align}\]

Substituting this value in equation \(\left( 2 \right)\), we get:

\[\begin{align}x\left[ { - \frac{{y'.y' + y.y''}}{{{a^2}}}} \right] + \frac{{yy'}}{{{a^2}}} &= 0\\\frac{x}{{{a^2}}}\left[ {y'.y' + y.y''} \right] - \frac{{yy'}}{{{a^2}}} &= 0\\ \Rightarrow \; x{{y'}^2} + xyy'' - yy' &= 0\\ \Rightarrow \; xyy'' + x{{y'}^2} - yy' &= 0\end{align}\]

Chapter 9 Ex.9.3 Question 10

Form the differential equation of the family of circles having centre on \(y\)-axis and radius \(3\) units.

 

Solution

 

Let centre of the circle on \(y\)-axis be \(\left( {0,b} \right)\).

\[\begin{align}&{x^2} + {\left( {y - b} \right)^2} = {3^2}\\& \Rightarrow \; {x^2} + {\left( {y - b} \right)^2} = 9 \qquad \ldots \left( 1 \right)\end{align}\]

Differentiating equation \(\left( 1 \right)\) with respect to \(x\), we get:

\[\begin{align}&2x + 2\left( {y - b} \right)y' = 0\\& \Rightarrow \; \left( {y - b} \right).y' = - x\\& \Rightarrow \; \left( {y - b} \right) = \frac{{ - x}}{{y'}}\end{align}\]

Substituting this value in equation \(\left( 1 \right)\), we get:

\[\begin{align}&{x^2} + {\left( {\frac{{ - x}}{{y'}}} \right)^2} = 9\\& \Rightarrow \; {x^2}\left[ {1 + \frac{1}{{{{\left( {y'} \right)}^2}}}} \right] = 9\\& \Rightarrow \; {x^2}\left[ {{{\left( {y'} \right)}^2} + 1} \right] = 9{\left( {y'} \right)^2}\\ &\Rightarrow \; \left( {{x^2} - 9} \right){\left( {y'} \right)^2} + {x^2} = 0\end{align}\]

Chapter 9 Ex.9.3 Question 11

Which of the following differential equations has \(y = {c_1}{e^x} + {c_2}{e^{ - x}}\) as the general solution?

(A) \(\frac{{{d^2}y}}{{d{x^2}}} + y = 0\)

(B) \(\frac{{{d^2}y}}{{d{x^2}}} + y = 0\)

(C) \(\frac{{{d^2}y}}{{d{x^2}}} + 1 = 0\)

(D) \(\frac{{{d^2}y}}{{d{x^2}}} - 1 = 0\)

 

Solution

 

\[\begin{align}&y = {c_1}{e^x} + {c_2}{e^{ - x}}\\ &\Rightarrow \; \frac{{dy}}{{dx}} = {c_1}{e^x} - {c_2}{e^{ - x}}\\ &\Rightarrow \; \frac{{{d^2}y}}{{d{x^2}}} = {c_1}{e^x} + {c_2}{e^{ - x}}\\& \Rightarrow \; \frac{{{d^2}y}}{{d{x^2}}} = y\\& \Rightarrow \; \frac{{{d^2}y}}{{d{x^2}}} - y = 0\end{align}\]

Thus, the correct option is B.

Chapter 9 Ex.9.3 Question 12

Which of the following differential equations has \(y = x\) as one of its particular solution?

(A) \(\frac{{{d^2}y}}{{d{x^2}}} - {x^2}\frac{{dy}}{{dx}} + xy = x\)

(B) \(\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}} + xy = x\)

(C) \(\frac{{{d^2}y}}{{d{x^2}}} - {x^2}\frac{{dy}}{{dx}} + xy = 0\)

(D) \(\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}} + xy = 0\)

 

Solution

 

\[\begin{align}\frac{{dy}}{{dx}} &= 1\\\frac{{{d^2}y}}{{d{x^2}}} &= 0\end{align}\]

Therefore,

\[\begin{align}\frac{{{d^2}y}}{{d{x^2}}} - {x^2}\frac{{dy}}{{dx}} + xy &= 0 - {x^2}.1 + x.x\\ &= - {x^2} + {x^2}\\& = 0\end{align}\]

Thus, the correct option is C.

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