Find the general solution of the differential equation x d y d x + 2 y = x 2 x ≠ 0 .

Given Differential Equation: x d y d x + 2 y = x 2 Dividing the above equation by x , d y d x + 2 x · y = x … … eq ( 1 ) Equation 1 is of the form d y d x + P y = Q Where, P = 2 x and Q = x Therefore, integrating factor is I . F . = e ∫ P d x = e ∫ 2 x d x = e 2 log x … … ∵ ∫ 1 x d x = log x = e log x 2 … … ∵ a log b = log b a = x 2 … … ∵ a log a b = b General solution is y · ( IF ) = ∫ Q · ( IF ) d x + c ⇒ y . ( x 2 ) = ∫ x . ( x 2 ) d x + c ⇒ x 2 y = ∫ x 3 d x + c ⇒ x 2 y = x 4 4 + c Therefore, the general solution of given differential equation is y = x 2 4 + c x 2 .

Show that the differential equation x 2 − y 2 d x + 2 x y d y = 0 is homogeneous and solve.

We have, x 2 − y 2 d x + 2 x y d y = 0 ⇒ 2 x y d y = − x 2 − y 2 d x ⇒ 2 x y d y = y 2 − x 2 d x ⇒ d y d x = y 2 − x 2 2 x y . . . 1 So, d y d x is a homogeneous differential equation of degree two. Solving d y d x by putting y = v x Differentiating y = v x , we get ⇒ d y d x = x · d v d x + v · d x d x ⇒ d y d x = x · d v d x + v Thus, x · d v d x + v = ( x v ) 2 − x 2 2 x ( x v ) ⇒ x · d v d x + v = x 2 v 2 − x 2 2 x 2 v ⇒ x · d v d x = x 2 v 2 − x 2 − 2 x 2 v 2 2 x 2 v ⇒ d v d x = − 1 x v 2 + 1 2 v ⇒ 2 v d v v 2 + 1 = − d x x Integrating both sides, we get: ∫ 2 v v 2 + 1 d v = ∫ − d x x ⇒ log 1 + v 2 = − log ⁡ | x | + c ⇒ log ⁡ v 2 + 1 + log ⁡ | x | = c ⇒ log ⁡ x v 2 + 1 = c ∵ log m + log n = log m · n ⇒ log ⁡ y x 2 + 1 x = c ⇒ log ⁡ y 2 + x 2 x = log ⁡ c ⇒ y 2 + x 2 = c x ⇒ x 2 + y 2 = c x is the general solution of the given differential equation.

Find the general solution of the differential equation d y d x = 1 + y 2 1 + x 2 .

Given differential equation is d y d x = 1 + y 2 1 + x 2 Now, separating the variables, then we get: d y 1 + y 2 = d x 1 + x 2 Applying integrating both sides, then we get: ∫ d y 1 + y 2 = ∫ d x 1 + x 2 ⇒ tan - 1 y = tan - 1 x + C Here, C = Constant of integration Hence, the general solution of given differential equation is tan - 1 y = tan - 1 x + C .

Find the particular solution of the differential equation x d y d x + 2 y = x 2 , x ≠ 0 , given that y = 0 when x = 1 .

Given differential equation is x d y d x + 2 y = x 2 ⇒ d y d x + 2 x y = x It is of the form d y d x + P y = Q , where P = 2 x , Q = x Integrating factor IF = e ∫ P d x = e ∫ 2 x d x = e 2 ln x = e ln x 2 ⇒ IF = x 2 General solution is given by y IF = ∫ Q × IF d x + c ⇒ y x 2 = ∫ x × x 2 d x + c ⇒ y x 2 = ∫ x 3 d x + c ⇒ y x 2 = x 4 4 + c . . . . . i On applying the given condition y = 0 , x = 1 , 0 = 1 4 + c ⇒ c = - 1 4 From equation i , y x 2 = x 4 4 - 1 4 ⇒ y = x 4 - 1 4 x 2

Solve the differential equation x - y d y - x + y d x = 0

Given differential equation x - y d y - x + y d x = 0 ⇒ x - y d y = x + y d x ⇒ d y d x = x + y x - y ⇒ d y d x = 1 + y x 1 - y x . . . . . i It is a homogeneous differential equation, because d y d x = f y x . Let y = v x Differentiate it with respect to x , ⇒ d y d x = v + x d v d x Now from equation i , v + x d v d x = 1 + v 1 - v ⇒ x d v d x = 1 + v 1 - v - v ⇒ x d v d x = 1 + v - v + v 2 1 - v ⇒ x d v d x = 1 + v 2 1 - v ⇒ 1 - v 1 + v 2 d v = d x x Integrate both sides, ⇒ ∫ 1 - v 1 + v 2 d v = ∫ d x x ⇒ ∫ 1 1 + v 2 d v - ∫ v 1 + v 2 d v = ln x + ln c . . . . . ii Let I = ∫ v 1 + v 2 d v Let 1 + v 2 = t , Differentiate with respect to x , ⇒ 2 v d v = d t Now, I = 1 2 ∫ 1 t d t ⇒ I = 1 2 ln t ⇒ I = 1 2 ln 1 + v 2 From equation ii , ⇒ ∫ 1 1 + v 2 d v - 1 2 ln 1 + v 2 = ln x + ln c ⇒ tan - 1 v - 1 2 ln 1 + v 2 = ln c x ⇒ tan - 1 y x - 1 2 ln 1 + y 2 x 2 = ln c x ⇒ tan - 1 y x - 1 2 ln x 2 + y 2 x 2 = ln c x ⇒ tan - 1 y x = ln c x + ln x 2 + y 2 x 2 ⇒ tan - 1 y x = ln c x × x 2 + y 2 x 2 ⇒ tan - 1 y x = ln c x 2 + y 2