Ncert Solutions For Class 12 Maths Ex 9.3

Ncert Solutions For Class 12 Maths Chapter 9 Ex 9.3

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

Solution:

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

Differentiate both the sides w.r.t x, we get:

\(\frac{1}{a}+\frac{1}{b}\frac{dy}{dx}=0\\ \\ \Rightarrow \frac{1}{a}+\frac{1}{b}y^{’}=0\)

Again, differentiate both the sides w.r.t x, we get:

\(0+\frac{1}{b}y^{’’}=0\\ \\ \Rightarrow \frac{1}{b}y^{’’}=0\\ \\ \Rightarrow y^{’’}=0\)

Hence, the required differential equation of the given curve is \(y^{’’}=0\).

Q.2: \(y^{2}=a(b^{2}-x^{2})\)

Solution:

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

Differentiate both the sides w.r.t x, we get:

\(2y\frac{dy}{dx}=a(-2x)\\ \\ \Rightarrow 2yy^{’}=-2ax\\ \\ \Rightarrow yy^{’}=-ax\) . . . . . . . . . . . . (1)

Again, differentiate both the sides w.r.t x, we get:

\(y^{’}.y^{’}+yy^{’’}=-a\\ \\ \Rightarrow (y^{’})^{2}+yy^{’’}=-a\)….(2)

Divide equation (2) by equation(1), we get:

\(\frac{(y^{’})^{2}+yy^{’’}}{yy^{’}}=\frac{-a}{-ax}\\ \\ \Rightarrow xyy^{’’}+x(y^{’})^{2}-yy^{’’}=0\)

This is the required differential equation of the given curve.

Q.3: \(y=ae^{3x}+be^{-2x}\)

 

Solution:

\(y=ae^{3x}+be^{-2x}\) ………….(1)

Differentiate both the sides w.r.t x, we get:

\(y^{’}=3ae^{3x}-2be^{-2x}\: \: \: \: \: ……(2)\)

Again, differentiate both the sides w.r.t x, we get:

\(y^{’’}=9ae^{3x}+4be^{-2x}\: \: \: \: \: ……(3)\)

Multiply equation(1) with 2 and then add it to equation (2), we get:

\((2ae^{3x}+2be^{-2x})+(3ae^{3x}-2be^{-2x})=2y+y^{’}\\ \\ \Rightarrow 5ae^{3x}=2y+y^{’}\\ \\ \Rightarrow ae^{3x}=\frac{2y+y^{’}}{5}\)

Now, multiplying equation (1) with 3 and subtracting equation (2) from it, we get:

\((3ae^{3x}+3be^{-2x})-(3ae^{3x}-2be^{-2x})=3y-y^{’}\\ \\ \Rightarrow 5be^{-2x}=3y-y^{’}\\ \\ \Rightarrow be^{-2x}=\frac{3y-y^{’}}{5}\)

Substituting the values of \(ae^{3x}\: and\: be^{-2x}\) in equation(3), we get:

\(y^{’’}=9\cdot \frac{(2y+y^{’})}{5}+4\cdot \frac{(3y-y^{’})}{5}\\ \\ \Rightarrow y^{’’}=\frac{18y+9y^{’}}{5}+\frac{12y-4y^{’}}{5}\\ \\ \Rightarrow y^{’’}=\frac{30y+5y^{’}}{5}\\ \\ \Rightarrow y^{’’}=6y+y^{’}\\ \\ \Rightarrow y^{’’}-y^{’}-6y=0\)

This is the required differential equation of the given curve.

Q.4: \(y=e^{2x}(a+bx)\)

Solution:

\(y=e^{2x}(a+bx)\) …..(1)

Differentiate both the sides w.r.t x, we get:

\(y^{’}=2e^{2x}(a+bx)+e^{2x}.b\\ \\ \Rightarrow y^{’}=e^{2x}(2a+2bx+b)\)…………(2)

Multiply equation (1) with 2 and then add it to equation (2), we get:

\(y^{’}-2y=e^{2x}(2a+2bx+b)-e^{2x}(2a+2bx)\\ \\ \Rightarrow y^{’}-2=be^{2x}\) ……(3)

Differentiate both the sides w.r.t x, we get:

\(y^{’’}k-2y^{’}=2be^{2x}\, \, \, \, \, \, \, \, \, \, …….(4)\)

Dividing equation (4) by equation (3), we get:

\(\frac{y^{’’}-2y^{’}}{y^{’}-2y}=2\\ \\ \Rightarrow y^{’’}-2y^{’}=2y^{’}-4y\\ \\ \Rightarrow y^{’’}-4y^{’}+4y=0\)

This is the required differential equation of the given curve.

 

 

Q.5: \(y=e^{x}(a\cos x+b\sin x)\)

 

Solution:

\(y=e^{x}(a\cos x+b\sin x)\) . . . . . . . . . . . . (1)

Differentiate both the sides w.r.t x, we get:

\(y^{’}=e^{x}(a\cos x+b\sin x)+e^{x}(-a\sin x+b\cos x)\\ \\ \Rightarrow y^{’}=e^{x}[(a+b)\cos x-(a-b)\sin x]\: \: \: \: \: …………(2)\)

Again, Differentiate both the sides w.r.t x, we get:

\(\Rightarrow y^{’’}=e^{x}[(a+b)\cos x-(a-b)\sin x]+e^{x}[-(a+b)\sin x-(a-b)\cos x]\\ \\ y^{’’}=e^{x}[2b\cos x-2a\sin x]\\ \\ y^{’’}=2e^{x}[b\cos x-a\sin x]\\ \\ \Rightarrow \frac{y^{’’}}{2}=e^{x}[b\cos x-a\sin x]\: \: \: \: \: \: \: ………..(3)\)

Adding equations (1) and (3), we get:

\(y+\frac{y^{’’}}{2}=e^{x}[(a+b)\cos x-(a-b)\sin x]\\ \\ \Rightarrow y+\frac{y^{’’}}{2}=y^{‘} \\ \\ \Rightarrow 2y+y^{’’}=2y^{’}\\ \\ \Rightarrow y^{’’}-2y^{’}+2y=0\)

This is the required differential equation of the given curve.

Q.6: Form the differential equation of the family of circles touching the y-axis at the origin.

Solution:

The centre of the circle touching the y-axis at origin lies on the x-axis.

Let (a, 0) be the centre of the circle.

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

Now, the equation of the circle with centre (a, 0) and radius (a) is

(x – a)2 + y2 = a2

i.e. x2 + y2 = 2ax . . . . . . . . . . . . . . . (1)

1

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

2x + 2yy’ = 2a

i.e. x + yy’ = a

Now, on substituting the value of a in equation (1), we get:

\(x^{2}+y^{2}=2(x+yy^{’})x\\ \\ \Rightarrow x^{2}+y^{2}=2x^{2}+2xyy^{’}\\ \\ \Rightarrow 2xyy^{’}+x^{2}=y^{2}\)

This is the required differential equation.

Q.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:

x2 = 4ay . . . . . . . . . . .(1)

2

Differentiate both the sides w.r.t x, we get:

2x = 4ay’ . . . . . . . . . . . . . . . . (2)

Dividing equation (2) by equation (1), we get:

\(\frac{2x}{x^{2}}=\frac{4ay^{’}}{4ay}\\ \\ \Rightarrow \frac{2}{x}=\frac{y^{’}}{y}\\ \\ \Rightarrow xy^{’}=2y\\ \\ \Rightarrow xy^{’}-2y=0\)

This is the required differential equation.

Q.8: Form the differential equation of the family of ellipses having foci on y-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:

\(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\: \: \: \: \: \: \: \: \: \: …..(1)\)

3

Differentiate both the sides w.r.t x, we get:

\(\frac{2x}{b^{2}}+\frac{2yy^{‘}}{b^{2}}=0\\ \\ \Rightarrow \frac{x}{b^{2}}+\frac{yy^{‘}}{a^{2}}=0\: \: \: \: \: \: \: \: (2)\)

Again, differentiate both the sides w.r.t x, we get:

\(\frac{1}{b^{2}}+\frac{y^{’}\cdot y^{’}+y.y^{’’}}{a^{2}}=0\\ \\ \Rightarrow \frac{1}{b^{2}}+\frac{1}{a^{2}}(y^{‘2}+yy^{’’})=0\\ \\ \Rightarrow \frac{1}{b^{2}}=-\frac{1}{a^{2}}(y^{’2}+yy^{’’})\)

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

\(x[-\frac{1}{a^{2}}((y^{’2})+yy^{’})]+\frac{yy^{’}}{a^{2}}=0\\ \\ \Rightarrow -x(y^{’})^{2}-xyy^{’’}+yy^{’}=0\\ \\ \Rightarrow xyy^{’’}+x(y^{‘})^{2}-yy^{’}=0\)

This is the required differential equation.

 

 

Q.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:

\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) …………(1)

4

Differentiate both the sides w.r.t x, we get:

\(\frac{2x}{a^{2}}-\frac{2yy^{’}}{b^{2}}=0\\ \\ \Rightarrow \frac{x}{a^{2}}-\frac{yy^{’}}{b^{2}}=0\) ……….. (2)

Again, differentiate both the sides w.r.t x, we get:

\(\frac{1}{a^{2}}-\frac{y^{’}\cdot y^{’}+yy^{’’}}{b^{2}}=0\\ \\ \Rightarrow \frac{1}{a^{2}}=\frac{1}{b^{2}}((y^{’})^{2}+yy^{’’})\)

Substituting the value of \(\frac{1}{a^{2}}\) in equation (2):

\(\frac{x}{b^{2}}((y^{’})^{2}+yy^{’’})-\frac{yy^{’}}{b^{2}}=0\\ \\ x(y^{’})^{2}+xyy^{’’}-yy^{’}=0\\ \\ \Rightarrow xyy^{’’}+x(y^{’})^{2}-yy^{’}=0\)

This is the required differential equation.

Q.10: Form the differential equation of the family of circles having centre on y-axis and radius 3 units.

Solution:

Let the center of the circle on y-axis be (0, b).

The differential equation of the family of circles with centre at (0, b) and radius 3 is as follows:

\(x^{2}+(y-b)^{2}=3^{2}\\ \\ \Rightarrow x^{2}+(y-b)^{2}=9\: \: \: \: \: …(1)\)

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

\(2x+2(y-b)\cdot y^{’}=0\\ \\ \Rightarrow (y-b)\cdot y^{’}=-x\\ \\ \Rightarrow y-b=\frac{-x}{y^{’}}\)

Substitute the value of (y – b) in equation (1), we get:

\(x^{2}+(\frac{-x}{y^{’}})^{2}=9\\ \\ \Rightarrow x^{2}[1+\frac{1}{(y^{’})^{2}}]=9\\ \\ \Rightarrow x^{2}((y^{‘})^{2}+1)=9(y^{’})^{2}\\ \\ \Rightarrow (x^{2}-9)(y^{’})^{2}+x^{2}=0\)

This is the required differential equation.

 

 

Q.11.Which of the following differential equations has \(y=c_{1}e^{x}+c_{2}e^{-x}\) as the general solution?

(i) \(\frac{d^{2}y}{dx^{2}}+y=0\)

(ii) \(\frac{d^{2}y}{dx^{2}}-y=0\)

(iii) \(\frac{d^{2}y}{dx^{2}}+1=0\)

(iv) \(\frac{d^{2}y}{dx^{2}}-1=0\)

Solution:

The given equation is:

\(y=c_{1}e^{x}+c_{2}e^{-x}\) . . . . . . . . . . . . . . . (1)

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

\(\frac{dy}{dx}=c_{1}e^{x}-c_{2}e^{-x}\)

Again, differentiate with respect to x, we get:

\(\frac{d^{2}y}{dx^{2}}=c_{1}e^{x}-c_{2}e^{-x}\\ \\ \Rightarrow \frac{d^{2}y}{dx^{2}}=y\\ \\ \Rightarrow \frac{d^{2}y}{dx^{2}}-y=0\)

This is the required differential equation of the given equation of curve.

Hence, the correct answer is (ii).

 

 

Q.12: Which of the following differential equation has y = x as one of its particular solution?

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

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

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

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

Solution:

The given equation of curve is y = x.

Differentiate with respect to x, we get:

\(\frac{dy}{dx}=1\: \: \: \: \: ….(1)\)

Again, differentiate with respect to x, we get:

\(\frac{d^{2}y}{dx^{2}}=0\: \: \: \: \: ….(2)\)

Now, on substituting the values of y:

\(\frac{d^{2}y}{dx^{2}}\), and \(\frac{dy}{dx}\) from equation (1) and (2) in each of the given alternatives, we find that only the differential equation given in alternative C is correct.

\(\frac{d^{2}y}{dx^{2}}-x^{2}\frac{dy}{dx}+xy=0-x^{2}\cdot 1+x\cdot x=-x^{2}+x^{2}=0\)

Hence, the correct answer is (iii).