Mathematics Solution Of Differential equations with variables separable Method For CBSE-NCERT
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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