Find the particular solution of the differential equation x dy dx = y - xtan y x , given that y = π 4 at x = 1 .

Given that, the differential equation is x dy dx = y - xtan y x . Also, given that y = π 4 at x = 1 The differential equation can be written as: d y d x = y x - tan y x Let y = v x ∴ dy dx = v + x d v dx ⇒ v + x d v dx = v - tan v ⇒ cot v d v = - 1 x dx Integrate both sides logsin v = - log | x | + log c ⇒ logsin y x = log c x ⇒ x · sin y x = c , Put y = π 4 and x = 1 ⇒ sin π 4 = c or c = 1 2 ∴ Particular solution is x · sin y x = 1 2

Verify that the function y = a cos x + b sin x , where a , b ∈ R is a solution of differential equation d 2 y d x 2 + y = 0

Consider y = a cos x + b sin x . . . i Differentiating ( i ) w.r.t x , we get d y d x = - a sin x + b cos x . . . i i Differentiating ( i i ) w.r.t. x , we get d 2 y d x 2 = - a cos x - b sin x d 2 y d x 2 + a cos x + b sin x = 0 By using equation ( i ) , we get d 2 y d x 2 + y = 0 . Thus, the function y = a cos x + b sin x , where a , b ∈ R is a solution of differential equation d 2 y d x 2 + y = 0 .

Find the order and the degree of the differential equation: x 2 d 2 y d x 2 = 1 + d y d x 2 4 .

Given differential equation is x 2 d 2 y d x 2 = 1 + d y d x 2 4 ⇒ x 2 d 2 y d x 2 1 = 1 + d y d x 2 4 We know that, The order of a differential equation is the order of the highest derivative occurring in the differential equation. The degree of a differential equation is the power of the highest order derivative when differential coefficients are made free from radicals and fractions. Hence, Order = 2 and Degree = 1 .

Find the particular solution of the differential equation d y d x + 2 y tan x = sin x , given that y = 0 when x = π 3 .

Given differentiation equation is d y d x + 2 y tan x = sin x . . . . i This equation is of the form d y d x + P y = Q . . . . i i , where P , and Q are either constants or function of x . Comparing equations i , and i i , we have P = 2 tan x , and Q = sin x Thus, the integrating factor for the given differential equation is given by I.F = e ∫ P d x = e ∫ 2 tan x d x = e 2 ∫ tan x d x = e 2 log sec x = e log sec 2 x = sec 2 x . Now, the solution for this differential equation is given by y × I . F = ∫ Q × I . F d x + C ⇒ y sec 2 x = ∫ sin x × sec 2 x d x + C ⇒ y sec 2 x = ∫ sin x cos x × sec x d x + C ∵ sec x = 1 cos x ⇒ y sec 2 x = ∫ tan x × sec x d x + C ⇒ y sec 2 x = sec x + C . . . . . i i i Now, when x = π 3 , y = 0 Substituting these values of x , and y in equation i i i , we have 0 = sec π 3 + C ⇒ C = - 2 Hence, from equation i i i , the particular solution of the given differential equation is y sec 2 x = sec x - 2

If y = x + 1 x , prove that 2 x d y d x = x - 1 x

We have, y = x + 1 x Differentiate with respect to x , ⇒ d y d x = d d x x + 1 x ⇒ d y d x = d d x x + d d x 1 x ⇒ d y d x = 1 2 x + - 1 2 x x ⇒ d y d x = 1 2 x - 1 2 x x ⇒ d y d x = x - 1 2 x x ⇒ 2 x d y d x = x - 1 x ⇒ 2 x d y d x = x x - 1 x ⇒ 2 x d y d x = x - 1 x

If y = 1 2 log 1 - cos 2 x 1 + cos 2 x , prove that d y d x = 2 cosec 2 x

We have, y = 1 2 log 1 - cos 2 x 1 + cos 2 x ⇒ y = 1 2 log 2 sin 2 x 2 cos 2 x ⇒ y = 1 2 log tan 2 x ⇒ y = 2 2 logtan x ⇒ y = logtan x Differentiate with respect to x , d y d x = logtan x = 1 tan x × d d x tan x = sec 2 x tan x = 1 cos 2 x × sin x cos x = 1 sin x cos x = 2 2 sin x cos x = 2 sin 2 x So, d y d x = 2 cosec 2 x

If y = log 3 x , x ≠ 0 , find d y d x ?

Given that y = log 3 x ⇒ d y d x = d d x log 3 x ⇒ d y d x = 1 3 x d d x 3 x ⇒ d y d x = 1 3 x 3 ⇒ d y d x = 1 x

If y = x - 1 log x - 1 - x + 1 log x + 1 , prove that d y d x = log x - 1 1 + x

We have, y = x - 1 log x - 1 - x + 1 log x + 1 Differentiating with respect to x , d y d x = d d x x - 1 log x - 1 - x + 1 log x + 1 = x - 1 d d x log x - 1 + log x - 1 d d x x - 1 - x + 1 d d x log x + 1 + log x + 1 d d x x + 1 = x - 1 × 1 x - 1 d d x x - 1 + log x - 1 × 1 - x + 1 × 1 x + 1 × d d x x + 1 + log x + 1 1 = 1 + log x - 1 - 1 + log x + 1 = log x - 1 - log x + 1 = log x - 1 x + 1 So, d y d x = log x - 1 x + 1