The differential equation y d y d x + x = k , ( k ∈ R ) represents:

Family of circles centered at x - axis

Family of circles centered at y - axis

Family of rectangular hyperbolas

Family of parabolas whose axis is x - axis

The differential equation y d y d x + x = k , ( k ∈ R ) represents:

Family of circles centered at x - axis

Solve the differential equation 1 + x 2 d y d x + 2 x y - 4 x 2 = 0 , subject to the initial condition y ( 0 ) = 0 .

Given differential equation 1 + x 2 d y d x + 2 x y - 4 x 2 = 0 , y ( 0 ) = 0 ⇒ 1 + x 2 d y d x + 2 x y = 4 x 2 Divide by 1 + x 2 on both sides, we get ⇒ d y d x + 2 x 1 + x 2 · y = 4 x 2 1 + x 2 It is of the form d y d x + P y = Q Here, P = 2 x 1 + x 2 , Q = 4 x 2 1 + x 2 Now, integrating factor I . F = e ∫ P d x ⇒ I . F = e ∫ 2 x 1 + x 2 d x ⇒ I . F = e log 1 + x 2 ∵ ∫ f ' ( x ) f x d x = log f x + C ⇒ I . F = 1 + x 2 ∴ The general solution is y × I . F = ∫ Q × I . F d x + C ⇒ y · 1 + x 2 = ∫ 4 x 2 1 + x 2 × 1 + x 2 d x + C ⇒ y · 1 + x 2 = 4 ∫ x 2 d x + C ⇒ y · 1 + x 2 = 4 x 3 3 + C → 1 Now put x = 0 , y = 0 in 1 , we get ⇒ C = 0 Put the value of C in 1 , we get ⇒ y · 1 + x 2 = 4 x 3 3 .

Solve the differential equation d y d x + y x = x 2 .

Given differential equation is, d y d x + y x = x 2 Which is linear differential equation of first order and is of the form, d y d x + P y = Q Where P = 1 x and Q = x 2 ∴ I . F = e ∫ P d x = e ∫ 1 x d x = e log x = x Multiply given equation by x , we get, x d y d x + y = x 3 ⇒ x y = x 4 4 + c ∴ y = x 3 4 + c x

Solve the differential equation d y d x = x log x

Given d y d x = x log x The above equation can be written as d y = x log x d x Integrate both sides ⇒ ∫ d y = ∫ x log x d x ⇒ y = log x ∫ x d x - ∫ ∫ x d x d d x log x d x (Using integration by part) ⇒ y = x 2 2 log x - ∫ x 2 2 1 x d x ⇒ y = x 2 2 log x - x 2 4 + C This is the required solution for the given differential equation.

Solve the differential equation: x d y - y d x = x 2 + y 2 d x , given that y = 0 when x = 1 .

Given the differential equation is x d y - y d x = x 2 + y 2 d x ⇒ x d y = x 2 + y 2 + y d x ⇒ d y d x = x 2 + y 2 + y x → 1 Here the numerator degree and denominator degree are having the same. It is a homogeneous differential equation Put y = v x and d y d x = v + x d v d x (Apply product rule in differentiation) Now put these values in 1 , we get ⇒ v + x d v d x = x 2 + v 2 x 2 + v x x = 1 + v 2 + v ⇒ x d v d x = 1 + v 2 ⇒ d v 1 + v 2 = d x x ⇒ ∫ d v 1 + v 2 = ∫ d x x (Apply integration) ⇒ log v + 1 + v 2 = log x + log C = log C x ⇒ v + 1 + v 2 = C x ⇒ y x + 1 + y 2 x 2 = C x Put y = v x ⇒ y + x 2 + y 2 = C x 2 → 2 When x = 1 , y = 0 (given), substitute these values in 2 , we get ⇒ 0 + 1 + 0 = C × ( 1 ) 2 ⇒ C = 1 ∴ y + x 2 + y 2 = x 2 is the required solution.