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NCERT Solutions for Class 12 Maths Chapter 9: Differential Equations - Exercise 9.3

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NCERT Solutions for Class 12 Maths Chapter 9 (Ex 9.3)

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.3 (Ex 9.3) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 9 Differential Equations Exercise 9.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.


Class:

NCERT Solutions for Class 12

Subject:

Class 12 Maths

Chapter Name:

Chapter 9 - Differential Equations

Exercise:

Exercise - 9.3

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2024-25

Medium:

English and Hindi

Available Materials:

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  • Exercise Wise

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  • Important Questions

  • Revision Notes

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Access NCERT Solution for Class 12 Maths Chapter 9 - Differential Equations

NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.3  are helpful for the students to improve their hold on the questions based on the sequences and series. All the questions of this exercise given in the pdf form on this page have been solved by subject experts. NCERT Solutions for Class 12 Maths Chapter Exercise 9.3 is based on the following topics:

  1. Formation of Differential Equations Whose Solution is Given

  2. Procedure to form a differential equation that will represent a given family of curves.

 

What Does Differential Equation Mean?

A differential equation is an equation that includes both a function and its derivatives. In other words, a differential equation is an equation making a statement that joins  the value of a quantity to the rate at which that quantity is changing.

1. $\dfrac{x}{a} + \dfrac{y}{b} = 1$

Ans: The given equation is,

$\dfrac{x}{a} + \dfrac{y}{b} = 1$…(i)

By differentiating each side of the equation (i) with respect to x,

$\dfrac{d}{{dx}}\left( {\dfrac{x}{a} + \dfrac{y}{b}} \right) = \dfrac{d}{{dx}}\left( 1 \right)$…(ii)

We have, $d\left( {x + y} \right) = d\left( x \right) + d\left( y \right)$

And $d\left( {cons\tan t} \right) = 0$

The equation (ii) becomes,

$\dfrac{d}{{dx}}\left( {\dfrac{x}{a}} \right) + \dfrac{d}{{dx}}\left( {\dfrac{y}{b}} \right) = 0$

By differentiating,

$\dfrac{1}{a} + \dfrac{1}{b}\dfrac{{dy}}{{dx}} = 0$

$\left[ {\because \dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}} \right]$

$\dfrac{1}{a} + \dfrac{1}{b}y' = 0$

$\left[ {\because \dfrac{{dy}}{{dx}} \equiv y'} \right]$

Again differentiating each sideof the above equation with respect to x,

$\dfrac{d}{{dx}}\left( {\dfrac{1}{a} + \dfrac{1}{b}y'} \right) = \dfrac{d}{{dx}}\left( 0 \right)$

This implies,

$\dfrac{d}{{dx}}\left( {\dfrac{1}{a}} \right) + \dfrac{d}{{dx}}\left( {\dfrac{1}{b}y'} \right) = \dfrac{d}{{dx}}\left( 0 \right)$

So,

$0 + \dfrac{1}{b}y'' = 0$

$\left[ {\because \dfrac{{dy'}}{{dx}} \equiv y''} \right]$

This implies, $\dfrac{1}{b}y'' = 0$

Multiplying $b$ with both sides,

$y'' = 0$

Hence, the required differential equation is $y'' = 0$.


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

Ans: The given equation is,

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

By differentiating each side of the equation (i) with respect to x,

$\dfrac{d}{{dx}}\left( {{y^2}} \right) = \dfrac{d}{{dx}}\left[ {a\left( {{b^2} - {x^2}} \right)} \right]$

This implies,

$2y\dfrac{d}{{dx}}\left( y \right) = a\left[ {\dfrac{d}{{dx}}\left( {{b^2}} \right) - \dfrac{d}{{dx}}\left( {{x^2}} \right)} \right]$

We have,

$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$

And

$d\left( {{\text{constant}}} \right) = 0$

So,

$2y\dfrac{{dy}}{{dx}} = a\left[ {0 - 2x} \right]$

This implies,

$2y\dfrac{{dy}}{{dx}} =  - 2ax$

Dividing both sides by 2,

$y\dfrac{{dy}}{{dx}} =  - ax$

We know, $\dfrac{{dy}}{{dx}} \equiv y'$

$yy' =  - ax$…(ii)

Again differentiating each side of the above equation with respect to $x$,

$\dfrac{d}{{dx}}\left( {yy'} \right) = \dfrac{d}{{dx}}\left( { - ax} \right)$…(iii)

We have,

$d\left( {x \cdot y} \right) = xd\left( y \right) + yd\left( x \right)$       [Multiplication property]

The equation (iii) becomes,

$y\dfrac{d}{{dx}}\left( {y'} \right) + y'\dfrac{d}{{dx}}\left( y \right) =  - a\dfrac{d}{{dx}}\left( x \right)$

We have ,$\dfrac{{dy'}}{{dx}} \equiv y''$

$yy'' + y'y' =  - a$

So,

$yy'' + {\left( {y'} \right)^2} =  - a$…(iv)

Dividing equation (iv) by equation (ii),

$\dfrac{{yy'' + {{\left( {y'} \right)}^2}}}{{yy'}} = \dfrac{{ - a}}{{ - ax}}$

By simplifying,

$\dfrac{{yy'' + {{\left( {y'} \right)}^2}}}{{yy'}} = \dfrac{1}{x}$

By cross multiplication,

$xyy'' + x{\left( {y'} \right)^2} = yy'$

This can be written as,

$xyy'' + x{\left( {y'} \right)^2} - yy' = 0$

Hence, the required differential equation is $xyy'' + x{\left( {y'} \right)^2} - yy' = 0$.


3. $y = a{e^{3x}} + b{e^{ - 2x}}$

Ans: The given equation is,

$y = a{e^{3x}} + b{e^{ - 2x}}$…(i)

By differentiating each side of the equation (i) with respect to $x$,

$\dfrac{d}{{dx}}\left( y \right) = \dfrac{d}{{dx}}\left( {a{e^{3x}} + b{e^{ - 2x}}} \right)$…(ii)

We have,$d\left( {x + y} \right) = d\left( x \right) + d\left( y \right)$

$\dfrac{{dy}}{{dx}} \equiv y'$

So, the equation (ii) becomes,

$y' = \dfrac{d}{{dx}}\left( {a{e^{3x}}} \right) + \dfrac{d}{{dx}}\left( {b{e^{ - 2x}}} \right)$

We know, $\dfrac{d}{{dx}}\left( {{e^{nx}}} \right) = n{e^{nx}}$

So,

$y' = 3a{e^{3x}} - 2b{e^{ - 2x}}$…(iii)

Again, differentiating each side of the above equation with respect to x,

$\dfrac{d}{{dx}}\left( {y'} \right) = \dfrac{d}{{dx}}\left( {3a{e^{3x}} - 2b{e^{ - 2x}}} \right)$

We have,$\dfrac{{dy'}}{{dx}} \equiv y''$

So,

$y'' = \dfrac{d}{{dx}}\left( {3a{e^{3x}}} \right) - \dfrac{d}{{dx}}\left( {2b{e^{ - 2x}}} \right)$

This implies,

$y'' = 3a \cdot 3{e^{3x}} - 2b \cdot ( - 2){e^{ - 2x}}$

By simplifying,

$y'' = 9a{e^{3x}} + 4b{e^{ - 2x}}$…(iv)

Now, multiplying (2) with the equation (i) ,

$2y = 2a{e^{3x}} + 2b{e^{ - 2x}}$…(v)

Adding equation (iii) and equation (v) we get,

$y' + 2y = 3a{e^{3x}} - 2b{e^{ - 2x}} + 2a{e^{3x}} + 2b{e^{ - 2x}}$

By simplifying,

$y' + 2y = 5a{e^{3x}}$

Dividing both sides by 5,

$a{e^{3x}} = \dfrac{{y' + 2y}}{5}$…(vi)

Now multiplying (3) with the equation (i) ,

$3y = 3a{e^{3x}} + 3b{e^{ - 2x}}$…(vii)

Subtracting equation (vii) from equation (iii) we get,

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

By simplifying,

$y' - 3y = 3a{e^{3x}} - 2b{e^{ - 2x}} - 3a{e^{3x}} - 3b{e^{ - 2x}}$

$y' - 3y =  - 5a{e^{ - 2x}}$

Dividing both sides by $\left( { - 5} \right)$,

$\dfrac{{y' - 3y}}{{ - 5}} = a{e^{ - 2x}}$

This implies,

$a{e^{ - 2x}} = \dfrac{{3y - y'}}{5}$…(viii)

Substituting the value of $a{e^{3x}}$and $a{e^{ - 2x}}$in the equation (iv) we get,

$y'' = 9 \cdot \dfrac{{y' + 2y}}{5} + 4 \cdot \dfrac{{3y - y'}}{5}$

 By simplifying, 

$y'' = \dfrac{{9y' + 18y}}{5} + \dfrac{{12y - 4y'}}{5}$

 Adding,

$y'' = \dfrac{{9y' + 18y + 12y - 4y'}}{5}$

Now by simplifying we get,

$y'' = \dfrac{{5y' + 30y}}{5}$

By simplifying we get,

$y'' = y' + 6y$

This implies,

$y'' - y' - 6y = 0$

Hence, the required differential equation is$y'' - y' - 6y = 0$ .


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

Ans: The given equation is

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

$y = a{e^{2x}} + bx{e^{2x}}$…(i)

By differentiating each side of the equation (i) with respect to $x$ ,

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

$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {a{e^{2x}}} \right) + \dfrac{d}{{dx}}\left( {bx{e^{2x}}} \right)$

$\dfrac{{dy}}{{dx}} = 2a{e^{2x}} + bx\dfrac{d}{{dx}}\left( {{e^{2x}}} \right) + b{e^{2x}}\dfrac{d}{{dx}}\left( x \right)$

$\dfrac{{dy}}{{dx}} = 2a{e^{2x}} + bx\left( {2{e^{2x}}} \right) + b{e^{2x}}\left( 1 \right)$

$\dfrac{{dy}}{{dx}} = 2a{e^{2x}} + 2bx{e^{2x}} + b{e^{2x}}$

We have, $\dfrac{{dy}}{{dx}} \equiv y'$

So,

$y' = 2a{e^{2x}} + 2bx{e^{2x}} + b{e^{2x}}$…(ii)

By multiplying equation (i) with 2,

$2y = 2a{e^{2x}} + 2bx{e^{2x}}$

Subtracting it from equation (ii) we get,

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

$y' - 2y = 2a{e^{2x}} + 2bx{e^{2x}} + b{e^{2x}} - 2a{e^{2x}} - 2bx{e^{2x}}$

$y' - 2y = b{e^{2x}}$…(iii)

Again differentiating each side of the equation with respect to $x$,

$\dfrac{d}{{dx}}\left( {y'} \right) - \dfrac{d}{{dx}}\left( {2y} \right) = \dfrac{d}{{dx}}\left( {b{e^{2x}}} \right)$

We have, $\dfrac{{dy'}}{{dx}} \equiv y''$

So,

$y'' - 2y' = 2b{e^{2x}}$…(iv)

Dividing the equation (iv) by equation (iii),

$\dfrac{{y'' - 2y'}}{{y' - 2y}} = \dfrac{{2b{e^{2x}}}}{{b{e^{2x}}}}$

By simplifying,

$\dfrac{{y'' - 2y'}}{{y' - 2y}} = 2$

By cross multiplication,

$y'' - 2y' = 2y' - 4y$

$y'' - 2y' - 2y' + 4y = 0$

 By simplifying,

$y'' - 4y' + 4y = 0$

Hence, the required differential equation is $y'' - 4y' + 4y = 0$.


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

Ans: The given equation is

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

$y = a{e^{3x}}\cos x + b{e^{3x}}\sin x$…(i)

By differentiating each side of the equation (i) with respect to $x$,

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

We have,

$d\left( {x + y} \right) = d\left( x \right) + d\left( y \right)$

So,

$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {a{e^{3x}}\cos x} \right) + \dfrac{d}{{dx}}\left( {b{e^{3x}}\sin x} \right)$

We have,

$d\left( {x \cdot y} \right) = xd\left( y \right) + yd\left( x \right)$

So,

$\dfrac{{dy}}{{dx}} = a{e^{3x}}\dfrac{d}{{dx}}\left( {\cos x} \right) + a\cos x\dfrac{d}{{dx}}\left( {{e^{3x}}} \right) + b{e^{3x}}\dfrac{d}{{dx}}\left( {\sin x} \right) + b\sin x\dfrac{d}{{dx}}\left( {{e^{3x}}} \right)$

$\dfrac{{dy}}{{dx}} = a{e^{3x}}\left( { - \sin x} \right) + a\cos x\left( {3{e^{3x}}} \right) + b{e^{3x}}\left( {\cos x} \right) + b\sin x\left( {3{e^{3x}}} \right)$

$\dfrac{{dy}}{{dx}} =  - a{e^{3x}}\sin x + 3a{e^{3x}}\cos x + b{e^{3x}}\cos x + 3b{e^{3x}}\sin x$

$\dfrac{{dy}}{{dx}} = \left( { - a + 3b} \right){e^{3x}}\sin x + \left( {3a + b} \right){e^{3x}}\cos x$

We have,

$\dfrac{{dy}}{{dx}} \equiv y'$

Hence,

$y' = \left( { - a + 3b} \right){e^{3x}}\sin x + \left( {3a + b} \right){e^{3x}}\cos x$…(ii)

Again, differentiating each side of the equation (ii) with respect to $x$,

$\dfrac{d}{{dx}}\left( {y'} \right) = \left( { - a + 3b} \right)\dfrac{d}{{dx}}\left( {{e^{3x}}\sin x} \right) + \left( {3a + b} \right)\dfrac{d}{{dx}}\left( {{e^{3x}}\cos x} \right)$

We have, $\dfrac{{dy'}}{{dx}} \equiv y''$' = $\left( { - a + 3b} \right)\left[ {{e^{3x}}\dfrac{d}{{dx}}\left( {\sin x} \right) + \sin x\dfrac{d}{{dx}}\left( {{e^{3x}}} \right)} \right] + \left( {3a + b} \right)\left[ {{e^{3x}}\dfrac{d}{{dx}}\left( {\cos x} \right) + \cos x\dfrac{d}{{dx}}\left( {{e^{3x}}} \right)} \right]$

$y'' = \left( { - a + 3b} \right)\left[ {{e^{3x}}\left( {\cos x} \right) + \sin x\left( {3{e^{3x}}} \right)} \right] + \left( {3a + b} \right)\left[ {{e^{3x}}\left( { - \sin x} \right) + \cos x\left( {3{e^{3x}}} \right)} \right]$

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

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

Combining like terms,

$y'' = \left[ {\left( { - a + 3b} \right) + 3\left( {3a + b} \right)} \right]{e^{3x}}\cos x + \left[ {3\left( { - a + 3b} \right) - \left( {3a + b} \right)} \right]{e^{3x}}\sin x$

$y'' = \left( { - a + 3b + 9a + 3b} \right){e^{3x}}\cos x + \left( { - 3a + 9b - 3a - b} \right){e^{3x}}\sin x$

$y'' = \left( {8a + 6b} \right){e^{3x}}\cos x + \left( { - 6a + 8b} \right){e^{3x}}\sin x$…(iv)

Multiplying (10) with the equation (i) we get,

$10y = 10a{e^{3x}}\cos x + 10b{e^{3x}}\sin x$…(v)

Now, adding equation (v) and equation (iv) we get,

$y'' + 10y = \left( {8a + 6b} \right){e^{3x}}\cos x + \left( { - 6a + 8b} \right){e^{3x}}\sin x + 10a{e^{3x}}\cos x + 10b{e^{3x}}\sin x$

$y'' + 10y = \left( {8a + 6b + 10a} \right){e^{3x}}\cos x + \left( { - 6a + 8b + 10b} \right){e^{3x}}\sin x$

$y'' + 10y = \left( {18a + 6b} \right){e^{3x}}\cos x + \left( { - 6a + 18b} \right){e^{3x}}\sin x$

Also, $y'' - 6y = 6\left( {3a + b} \right){e^{3x}}\cos x + 2\left( { - a + 3b} \right){e^{3x}}\sin x$

This implies,

$y'' + 10y = 6\left[ {\left( {3a + b} \right){e^{3x}}\cos x + \left( { - a + 3b} \right){e^{3x}}\sin x} \right]$

Substituting equation (ii) we get,

$y'' + 10y = 6y'$

So,

$y'' - 6y' + 10y = 0$

Hence, the required differential equation is $y'' - 6y' + 10y = 0$. 


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

Ans: The centre of the circle, touching the y-axis at origin $\left( {0,0} \right)$ lies on the $x$-axis.

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

Since, it touches the y-axis at origin$\left( {0,0} \right)$

So, the radius of the circle is $c$.

Now, the equation of the circle with centre $\left( {c,0} \right)$and radius c is

${\left( {x - c} \right)^2} + {y^2} = {c^2}$

By expanding,

${x^2} - 2cx + {c^2} + {y^2} = {c^2}$

Subtracting ${c^2}$from both sides,

${x^2} - 2cx + {y^2} = 0$ …(i)

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By differentiating each side of the equation (i) with respect to $x$,

$2x + 2yy' = 2c$

Dividing by 2,

$x + yy' = c$

Substituting value of $c$ in the equation (i),

${x^2} - 2\left( {x + yy'} \right)x + {y^2} = 0$

${x^2} - 2{x^2} - 2xyy' + {y^2} = 0$

$ - {x^2} - 2xyy' + {y^2} = 0$

Multiplying $\left( { - 1} \right)$ to both sides,

${x^2} + 2xyy' - {y^2} = 0$

$2xyy' + {x^2} = {y^2}$

Hence, the required differential equation is $2xyy' + {x^2} = {y^2}$ .


7. Form the differential equation of the family of parabolas having vertex at origin and axis along positive $y$-axis.

Ans: The equation of the parabola having the vertex at origin$\left( {0,0} \right)$ and axis along the positive $y$-axis is

${x^2} = 4by$…(i)

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By differentiating each side of the equation (i) with respect to $x$,

$2x = 4ay'$…(ii)

Now, dividing equation (ii) by equation (i),

$\dfrac{{2x}}{{{x^2}}} = \dfrac{{4ay'}}{{4ay}}$

By simplifying,

$\dfrac{2}{x} = \dfrac{{y'}}{y}$

By cross multiplication,

$xy' = 2y$

This implies,

$xy' - 2y = 0$

Hence, the required differential equation is $xy' - 2y = 0$ .


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

Ans: The equation of the ellipse having foci on the $y$-axis and centre at origin$\left( {0,0} \right)$ is,

$\dfrac{{{x^2}}}{{{b^2}}} + \dfrac{{{y^2}}}{{{a^2}}} = 1$…(i)

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By differentiating each side of the equation (i) with respect to $x$,

$\dfrac{{2x}}{{{b^2}}} + \dfrac{{2yy'}}{{{a^2}}} = 0$…(ii)

Dividing by 2,

$\dfrac{x}{{{b^2}}} + \dfrac{{yy'}}{{{a^2}}} = 0$…(ii)

Again differentiating each side of the equation (ii) with respect to $x$,

$\dfrac{1}{{{b^2}}} + \dfrac{{y \cdot y'' + y' \cdot y'}}{{{a^2}}} = 0$

This implies,

$\dfrac{1}{{{b^2}}} + \dfrac{1}{{{a^2}}}\left( {yy'' + {{y'}^2}} \right) = 0$

So,

$\dfrac{1}{{{b^2}}} =  - \dfrac{1}{{{a^2}}}\left( {yy'' + {{y'}^2}} \right)$

Substituting this value in the equation (ii),

$x\left[ { - \dfrac{1}{{{a^2}}}\left( {yy'' + {{y'}^2}} \right)} \right] + \dfrac{{yy'}}{{{a^2}}} = 0$

$\dfrac{1}{{{a^2}}}\left[ { - x\left( {yy'' + {{y'}^2}} \right) + yy'} \right] = 0$

Multiplying ${a^2}$ to both sides,

$ - x\left( {yy'' + {{y'}^2}} \right) + yy' = 0$

$ - xyy'' - x{y'^2} + yy' = 0$

Multiplying $\left( { - 1} \right)$ to both sides,

$xyy'' + x{y'^2} - yy' = 0$

Hence, the required differential equation is  $xyy'' + x{y'^2} - yy' = 0$ .


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

Ans: The equation of the hyperbola with the centre at origin and foci along the x-axis is,

$\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$

(Image will be Uploaded Soon)

By differentiating each side of the equation (i) with respect to $x$,

$\dfrac{{2x}}{{{a^2}}} - \dfrac{{2yy'}}{{{b^2}}} = 0$…(ii)

Dividing by 2,

$\dfrac{x}{{{a^2}}} - \dfrac{{yy'}}{{{b^2}}} = 0$…(ii)

Again, differentiating each side of the equation (ii) with respect to $x$,

$\dfrac{1}{{{a^2}}} - \dfrac{{y \cdot y'' + y' \cdot y'}}{{{b^2}}} = 0$

This implies,

$\dfrac{1}{{{a^2}}} - \dfrac{1}{{{b^2}}}\left( {yy'' + {{y'}^2}} \right) = 0$

So,

$\dfrac{1}{{{a^2}}} = \dfrac{1}{{{b^2}}}\left( {yy'' + {{y'}^2}} \right)$

Substituting this value in the equation (ii),

$x\left[ {\dfrac{1}{{{b^2}}}\left( {yy'' + {{y'}^2}} \right)} \right] - \dfrac{{yy'}}{{{b^2}}} = 0$

$\dfrac{1}{{{b^2}}}\left[ {x\left( {yy'' + {{y'}^2}} \right) - yy'} \right] = 0$

Multiplying ${b^2}$ to both sides,

$x\left( {yy'' + {{y'}^2}} \right) - yy' = 0$

$xyy'' + x{y'^2} - yy' = 0$

Hence, the required differential equation is  $xyy'' + x{y'^2} - yy' = 0$ .


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

Ans: Let the centre of the circle on $y$-axis is $\left( {0,d} \right)$.

The differential equation of the circle with centre at $\left( {0,d} \right)$ and radius 3 units is,

${x^2} + {\left( {y - d} \right)^2} = {3^2}$

This implies,

${x^2} + {\left( {y - d} \right)^2} = 9$…(i)

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By differentiating each side of the equation (i) with respect to $x$,

$2x + 2\left( {y - d} \right)y' = 0$

Subtracting $2x$ from both sides,

$2\left( {y - d} \right)y' =  - 2x$

$\left( {y - d} \right)y' =  - x$

Dividing by $y'$ , 

$\left( {y - d} \right) =  - \dfrac{x}{{y'}}$

Now substituting this value in the equation (i),

${x^2} + {\left( { - \dfrac{x}{{y'}}} \right)^2} = 9$

${x^2} + \dfrac{{{x^2}}}{{{{y'}^2}}} = 9$

${x^2}\left[ {1 + \dfrac{1}{{{{y'}^2}}}} \right] = 9$

Adding,

${x^2}\left[ {\dfrac{{{{y'}^2} + 1}}{{{{y'}^2}}}} \right] = 9$

By cross multiplication,

${x^2}{y'^2} + {x^2} = 9{y'^2}$

${x^2}{y'^2} + {x^2} - 9{y'^2} = 0$

Combining like terms,

$\left( {{x^2} - 9} \right){y'^2} + {x^2} = 0$

Hence, the required differential equation is  $\left( {{x^2} - 9} \right){y'^2} + {x^2} = 0$ .


11. Which of the following differential equations has$y = {c_1}{e^x} + {c_2}{e^{ - x}}$ as the general solution?

  1. $\dfrac{{{d^2}y}}{{d{x^2}}} + y = 0$

  2. $\dfrac{{{d^2}y}}{{d{x^2}}} - y = 0$

  3. $\dfrac{{{d^2}y}}{{d{x^2}}} + 1 = 0$

  4. $\dfrac{{{d^2}y}}{{d{x^2}}} - 1 = 0$

Ans: The equation is given by,

$y = {c_1}{e^x} + {c_2}{e^{ - x}}$…(i)

By differentiating each side of the equation (i) with respect to $x$,

$\dfrac{{dy}}{{dx}} = {c_1}{e^x} - {c_2}{e^{ - x}}$…(ii)

Again, differentiating each side of the equation (ii) with respect to $x$,

$\dfrac{{{d^2}y}}{{d{x^2}}} = {c_1}{e^x} + {c_2}{e^{ - x}}$…(iii)

Now from equation (i) and equation (iii),

$\dfrac{{{d^2}y}}{{d{x^2}}} = y$

Thus,

$\dfrac{{{d^2}y}}{{d{x^2}}} - y = 0$

Hence, the required differential equation is $\dfrac{{{d^2}y}}{{d{x^2}}} - y = 0$.

So, the correct option is B.


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

  1. $\dfrac{{{d^2}y}}{{d{x^2}}} - {x^2}\dfrac{{dy}}{{dx}} + xy = x$

  2. $\dfrac{{{d^2}y}}{{d{x^2}}} - x\dfrac{{dy}}{{dx}} + xy = x$

  3. $\dfrac{{{d^2}y}}{{d{x^2}}} - {x^2}\dfrac{{dy}}{{dx}} + xy = 0$

  4. $\dfrac{{{d^2}y}}{{d{x^2}}} - x\dfrac{{dy}}{{dx}} + xy = 0$

Ans: The equation is given by,

$y = x$…(i)

By differentiating each side of the equation (i) with respect to $x$,

$\dfrac{{dy}}{{dx}} = 1$…(ii)

Again, differentiating each side of the equation (ii) with respect to $x$,

$\dfrac{{{d^2}y}}{{d{x^2}}} = 0$…(iii)

Now substituting the all values $y,\dfrac{{dy}}{{dx}}$ and $\dfrac{{{d^2}y}}{{d{x^2}}}$  from equations (i),(ii) and (ii) in the left hand side of each given options,

For option (A),

$\dfrac{{{d^2}y}}{{d{x^2}}} - {x^2}\dfrac{{dy}}{{dx}} + xy = 0 - {x^2} \cdot 1 + x \cdot x$

$\dfrac{{{d^2}y}}{{d{x^2}}} - {x^2}\dfrac{{dy}}{{dx}} + xy = 0$

So, option (A) is incorrect.

For option (B),

$\dfrac{{{d^2}y}}{{d{x^2}}} - x\dfrac{{dy}}{{dx}} + xy = 0 - x \cdot 1 + x \cdot x$

$\dfrac{{{d^2}y}}{{d{x^2}}} - x\dfrac{{dy}}{{dx}} + xy =  - x + {x^2}$

So, option (B) is incorrect.

For option (C),

$\dfrac{{{d^2}y}}{{d{x^2}}} - {x^2}\dfrac{{dy}}{{dx}} + xy = 0 - {x^2} \cdot 1 + x \cdot x$

$\dfrac{{{d^2}y}}{{d{x^2}}} - {x^2}\dfrac{{dy}}{{dx}} + xy = 0$

So, option (C) is correct.

Hence, $\dfrac{{{d^2}y}}{{d{x^2}}} - {x^2}\dfrac{{dy}}{{dx}} + xy = 0$ is the required differential equation.


NCERT Solutions for Class 12 Maths PDF Download

 

NCERT Solution Class 12 Maths of Chapter 9 All Exercises

Chapter 9 - Differential Equations Exercises in PDF Format

Exercise 9.1

12 Questions & Solutions (10 Short Answers, 2 MCQs)

Exercise 9.2

12 Questions & Solutions (10 Short Answers, 2 MCQs)

Exercise 9.3

12 Questions & Solutions (5 Short Answers, 5 Long Answers, 2 MCQs)

Exercise 9.4

23 Questions & Solutions (10 Short Answers, 12 Long Answers, 1 MCQs)

Exercise 9.5

17 Questions & Solutions (15 Short Answers, 2 MCQs)

Exercise 9.6 

19 Questions & Solutions (15 Short Answers, 2 Long Answers, 2 MCQs)


NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.3

Opting for the NCERT solutions for Ex 9.3 Class 12 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 9.3 Class 12 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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