`color{red} ♦ ` Differential equations with variables separable

In this section we shall discuss three methods of solving first order first degree differential equations.

● A first order-first degree differential equation is of the form

`color{red} {(dy)/(dx) = F(x,y)}` .........................(1)

● If `F (x, y)` can be expressed as a product `g (x) h(y),` where, `g(x)` is a function of `x` and `h(y)` is a function of `y,` then the differential equation (1) is said to be of variable separable type. The differential equation (1) then has the form

`color{red} {(dy)/(dx) = h (y)* g(x)}` ..............................(2)

● If `h (y) ≠ 0,` separating the variables, (2) can be rewritten as

`color{red} {1/(h(y) ) dy = g(x) dx}` ...........................(3)

`=>` Integrating both sides of (3), we get

`color{red} {int(1/(h(y) ) dy )= int(g(x) dx)}` ........................(4)

Thus, (4) provides the solutions of given differential equation in the form

`H(y) = G(x) + C`

`=>` Here, `H (y)` and `G (x)` are the anti derivatives of `1/(h(y) )` and `g (x)` respectively and `C` is the arbitrary constant.

`color{red} {(dy)/(dx) = F(x,y)}` .........................(1)

● If `F (x, y)` can be expressed as a product `g (x) h(y),` where, `g(x)` is a function of `x` and `h(y)` is a function of `y,` then the differential equation (1) is said to be of variable separable type. The differential equation (1) then has the form

`color{red} {(dy)/(dx) = h (y)* g(x)}` ..............................(2)

● If `h (y) ≠ 0,` separating the variables, (2) can be rewritten as

`color{red} {1/(h(y) ) dy = g(x) dx}` ...........................(3)

`=>` Integrating both sides of (3), we get

`color{red} {int(1/(h(y) ) dy )= int(g(x) dx)}` ........................(4)

Thus, (4) provides the solutions of given differential equation in the form

`H(y) = G(x) + C`

`=>` Here, `H (y)` and `G (x)` are the anti derivatives of `1/(h(y) )` and `g (x)` respectively and `C` is the arbitrary constant.

Q 3117434380

Find the general solution of the differential equation `(dy)/(dx) - (x+1)/ (2 -y) , \ \ \ \ \ \ (y!=2)`

Class 12 Chapter 9 Example 9

Class 12 Chapter 9 Example 9

We have

`(dy)/(dx) = (x+1)/(2-y)` ..........................(1)

Separating the variables in equation (1), we get

`(2 – y) dy = (x + 1) dx` .................. (2)

Integrating both sides of equation (2), we get

`int(2 -y) dy = int (x +1) dx`

or `2y - y^2/2 = x^2/2 + x +C_1`

or ` x^2 +y^2 +2x -4y +2 C_1 =0`

or `x^2 +y^2 +2x - 4y +C =0` where `C= 2 C_1`

which is the general solution of equation (1).

Q 3137434382

Find the general solution of the differential equation `(dy)/(dx) = (1 +y^2)//(1 +x^2)`

Class 12 Chapter 9 Example 10

Class 12 Chapter 9 Example 10

Since `1 + y^2 ≠ 0`, therefore separating the variables, the given differential equation can be written as

`(dy)/(1 +y^2) = (dx)/(1 +x^2)` ..............(1)

Integrating both sides of equation (1), we get

`int(dy)/(1 +y^2) = int (dx)/(1 +x^2)`

or `tan^-1 y = tan^-1x + C`

which is the general solution of equation (1).

Q 3177434386

Find the particular solution of the differential equation `(dy)/(dx) = - 4xy^2` given that y = 1, when x = 0.

Class 12 Chapter 9 Example 11

Class 12 Chapter 9 Example 11

If y ≠ 0, the given differential equation can be written as

`(dy)/y^2 = -4x dx` .....................(1)

Integrating both sides of equation (1), we get

`int(dy)/y^2 = - 4 int x dx`

or ` -1/y = - 2x^2 +C`

or ` y = 1/(2x^2 -C)` ......................(2)

Substituting y = 1 and x = 0 in equation (2), we get, C = – 1.

Now substituting the value of C in equation (2), we get the particular solution of the given differential equation as `y = 1/(2x^2 +1)`

Q 3107434388

Find the equation of the curve passing through the point (1, 1) whose differential equation is `x dy = (2x^2 + 1) dx \ \ \ \ (x ≠ 0)`.

Class 12 Chapter 9 Example 12

Class 12 Chapter 9 Example 12

The given differential equation can be expressed as

`dy^** = ( (2x^2 +1)/x) dx^**`

or `dy = (2 x + 1/x) dx` .......................(1)

Integrating both sides of equation (1), we get

`int dy = int (2 x +1/x) dx`

or `y =x^2 + log |x| +C` ............(2)

Equation (2) represents the family of solution curves of the given differential equation but we are interested in finding the equation of a particular member of the family which passes through the point (1, 1). Therefore substituting x = 1, y = 1 in equation (2), we

get C = 0.

Now substituting the value of C in equation (2) we get the equation of the required curve as `y = x^2 + log | x |`.

Q 3147534483

Find the equation of a curve passing through the point (–2, 3), given that the slope of the tangent to the curve at any point (x, y) is `(2x)/y^2`

Class 12 Chapter 9 Example 13

Class 12 Chapter 9 Example 13

We know that the slope of the tangent to a curve is given by `(dy)/(dx)`

so, `(dy)/(dx) = (2x)/y^2` ...................(1)

Separating the variables, equation (1) can be written as

`y^2 dy = 2x dx` ..................... (2)

Integrating both sides of equation (2), we get

`inty^2dy = int 2x dx`

or `y^2//3 = x^2 +C` ............................(3)

Substituting x = –2, y = 3 in equation (3), we get C = 5.

Substituting the value of C in equation (3), we get the equation of the required curve as

`y^3/3 =x^2 +5 ` ot `y = (3x^2 +15)^(1/3)`

Q 3177534486

In a bank, principal increases continuously at the rate of 5% per year. In how many years Rs 1000 double itself?

Class 12 Chapter 9 Example 14

Class 12 Chapter 9 Example 14

Let P be the principal at any time t. According to the given problem,

`(dp)/(dt) = (5/100) xx P`

or `(dp)/(dt) = P/20` ..............(1)

separating the variables in equation (1), we get

`(dp)/P = (dt)/20` ..........................(2)

Integrating both sides of equation (2), we get

`log P = t/20 +C_1`

or ` P =e^(t/20) * e^(C_1)`

or `P = Ce^(t/20)` (where `e^(C_1) =C`) ...............................(3)

Now P = 1000, when t = 0

Substituting the values of P and t in (3), we get C = 1000. Therefore, equation (3), gives

`P= 1000e^(t/20)`

Let t years be the time required to double the principal. Then

`2000 = 1000 e^(t/20) `

`=> t =20 log_e2`