NCERT Solutions for Class 12-commerce Maths CBSE Chapter 9: Differential Equations

Differential Equations Exercise Ex. 9.1

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Differential Equations Exercise Ex. 9.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Differential Equations Exercise Ex. 9.3

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Let x and y be the x-coordinate and y-coordinate of the point on the curve respectively.

We know that the slope of a tangent to the curve in the coordinate axes is given by the relation,

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

.

.

Solution 23

Differential Equations Exercise Ex. 9.4

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Differential Equations Exercise Ex. 9.5

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Differential Equations Exercise Misc. Ex.

Solution 1

Solution 2

$begin mathsize 14px style left parenthesis straight i right parenthesis space x y equals a e to the power of x plus b e to the power of negative x end exponent plus x squared Differentiating space both space sides space straight w. straight r. straight t space x space we space get fraction numerator d left parenthesis x y right parenthesis over denominator d x end fraction equals a fraction numerator d over denominator d x end fraction open parentheses e to the power of x close parentheses plus b fraction numerator d over denominator d x end fraction open parentheses e to the power of negative x end exponent close parentheses plus fraction numerator d over denominator d x end fraction left parenthesis 2 x right parenthesis rightwards double arrow x fraction numerator d y over denominator d x end fraction plus y equals a e to the power of x minus b e to the power of negative x end exponent plus 2 x Again comma space differentiating space both space the space sides space straight w. straight r. straight t. space x comma space we space get x fraction numerator d squared y over denominator d x squared end fraction plus fraction numerator d y over denominator d x end fraction plus fraction numerator d y over denominator d x end fraction equals a fraction numerator d over denominator d x end fraction left parenthesis e to the power of x right parenthesis minus b fraction numerator d over denominator d x end fraction left parenthesis e to the power of negative x end exponent right parenthesis plus 2 fraction numerator d over denominator d x end fraction left parenthesis x right parenthesis rightwards double arrow x fraction numerator d squared y over denominator d x squared end fraction plus 2 fraction numerator d y over denominator d x end fraction equals a e to the power of x plus b e to the power of negative x end exponent plus 2 rightwards double arrow x fraction numerator d squared y over denominator d x squared end fraction plus 2 fraction numerator d y over denominator d x end fraction equals x y minus x squared plus 2 rightwards double arrow x fraction numerator d squared y over denominator d x squared end fraction plus 2 fraction numerator d y over denominator d x end fraction minus x y plus x squared minus 2 equals 0 Hence comma space the space given space function space is space straight a space solution space of space the space corresponding space differential space equation. end style$