Find a particular solution satisfying the given condition for differential equations. ​dy/dx + y tan x = 2x + x^2 tan x, given that y = 1 when x = 0.

Question

Find a particular solution satisfying the given condition for \(\frac{dy}{dx}\)+ y tan x = 2x + x² tan x, given that y = 1 when x = 0 differential equations.

dy/dx + y tan x = 2x + x² tan x, given that y = 1 when x = 0.

Answer 1

Given Differential Equation :

\(\frac{dy}{dx}\)+ y tan x = 2x + x² tan x

Formula :

i) \(\int\) tan x dx log |sec x|

ii) a^{log_a b} = b

iii) \(\int\) u. v dx = u. \(\int\) v dx - \(\int\) \((\frac{du}{dx}. \int v\,dx)\) dx

iv) \(\int\) sec x. tan x dx = sec x

v) \(\frac{dy}{dx}\) (xⁿ) nx^n-1

vi) General solution :

For the differential equation in the form of

\(\frac{dy}{dx} \, + Py \, =Q\)

Where, integrating factor,

I.F. = \(e^{\int p\, dx}\)

Given differential equation is

\(\frac{dy}{dx}\) + y tan x = 2x + x² tan x ……eq(1)

Equation (1) is of the form

\(\frac{dy}{dx} \, + Py \, =Q\)

Where, P = tan x and Q = 2x +x² tan x

Therefore, integrating factor is

General solution is

Substituting I in eq(2),

Multiplying above equation by cos x,

∴ y =x² + c. (cos x)

Therefore, general solution is

y = x² + c. (cos x)

For particular solution put y=1 and x=0 in above equation,

∴ 1 = 0 + c

∴ c = 1

Substituting c in general solution,

∴ y = x² + cos x

Therefore, particular solution is

y = x² + cos x