Mathematics Solution Of Differential equations with variables separable Method For CBSE-NCERT
Click for Only Video

### Topic Covered

color{red} ♦  Differential equations with variables separable

### Methods of Solving First Order, First Degree Differential Equations

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

### Differential equations with variables separable

● 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.
Q 3117434380

Find the general solution of the differential equation (dy)/(dx) - (x+1)/ (2 -y) , \ \ \ \ \ \ (y!=2)
Class 12 Chapter 9 Example 9
Solution:

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
Solution:

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
Solution:

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
Solution:

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
Solution:

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
Solution:

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