Question

For the differential equation find a particular solution satisfying the given condition:

`(x^3 + x^2 + x + 1) dy/dx = 2x^2 + x; y = 1` When x = 0

Answer 1

We have, `(x^3 + x^2 + x + 1) dy/dx =2 x^x + x`

⇒ `dy/dx = (2x^3 + x)/(x^3 + x^2 + x + 1)`

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

Integrating (1) both sides, we get

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

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

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

Now, let `(2x^2 + x)/((x^2 +1) (x + 1)) = A/(x + 1) + (Bx + C)/ (x^2 + 1)`

⇒ 2x² + x = A (x² + 1) + (Bx + C) (x + 1)

Comparing coefficients of x, we get

1 = B + C               ....(2)

Comparing constant terms, we get

0 = A + C                

⇒ C = - A                  ...(3)

Comparing coefficients of x², we get

2 = A + B                   .....(4)

From (2) & (3), we get

-A + B = 1

Solving (4) & (5), we get 2B = 3

⇒ `B = 3/2` substituting the value of B in (2) & (5), we get

`A = 1/2` and `C = -1/2`

∴ `y = int[(1/2)/(x + 1) + (3/2 x - 1/2)/(x^2 + 1)]  dx`

⇒ `y = 1/2 log (x + 1) + 3/4 int (2x)/ (x^2 + 1)  dx - 1/2 int dx/(x^2 + 1)`

⇒ `y = 1/2 log (x + 1) + 3/4 log (x^2 + 1) - 1/2 tan^-1 x + C`

When x = 0, y = 1

⇒ `1 = 1/2 log 1 + 3/4 log 1 - 1/2 tan^-1 (0) + C`

⇒ C = 1

Hence, a particular solution is

`y = 1/2 log (x + 1) + 3/4 log (x^2 + 1) - 1/2 tan^-1 x +1`

⇒ `y = 1/4 log [(x + 1)^2 (x^2 + 1)^3] - 1/2 tan^-1 x + 1`