Solve the differential equation d y d x + 1 = e x + y .

Given differential equation d y d x + 1 = e x + y .....(i) Put x + y = t ⇒ 1 + d y d x = d t d x Eq(i) becomes, d t d x = e t ⇒ e - t d t = d x Now integrating both side, we get - e - t = x + C ⇒ - 1 e x + y = x + C ∵ t = x + y ⇒ - 1 = ( x + C ) e x + y ⇒ ( x + C ) e x + y + 1 = 0 Therefore, the required solution is ( x + C ) e x + y + 1 = 0 .

Solve the differential equation d y d x + 1 = e x + y .

Given differential equation d y d x + 1 = e x + y .....(i) Put x + y = t ⇒ 1 + d y d x = d t d x Eq(i) becomes, d t d x = e t ⇒ e - t d t = d x Now integrating both side, we get - e - t = x + C ⇒ - 1 e x + y = x + C ∵ t = x + y ⇒ - 1 = ( x + C ) e x + y ⇒ ( x + C ) e x + y + 1 = 0 Therefore, the required solution is ( x + C ) e x + y + 1 = 0 .

Solve the differential equation d y d x = 1 + x + y 2 + x y 2 , when y = 0 , x = 0 .

Given that, d y d x = 1 + x + y 2 + x y 2 ⇒ d y d x = 1 + x + y 2 1 + x ⇒ d y d x = 1 + y 2 1 + x ⇒ d y 1 + y 2 = 1 + x d x On integrating both sides, we get tan - 1 y = x + x 2 2 + K ......(i) When y = 0 and x = 0 , then substituting these values in Eq(i), we get tan - 1 ( 0 ) = 0 + 0 + K ⇒ K = 0 ⇒ tan - 1 y = x + x 2 2 ⇒ y = tan x + x 2 2 Therefore, the required solution is y = tan x + x 2 2 .

Find the general solution of x + 2 y 3 d y d x = y .

Given that, x + 2 y 3 d y d x = y ⇒ y · d x d y = x + 2 y 3 ⇒ d x d y = x y + 2 y 2 [Dividing throughout by y ] ⇒ d x d y - x y = 2 y 2 Which is a linear differential equation. On comparing it with d x d y + P x = Q , we get P = - 1 y , Q = 2 y 2 IF = e ∫ - 1 y d y = e - ∫ 1 y d y ∴ IF = e - log y = 1 y The general solution is x · IF = ∫ Q · IF d y + C ⇒ x · 1 y = ∫ 2 y 2 · 1 y d y + C ⇒ x y = 2 y 2 2 + C ⇒ x y = y 2 + C ⇒ x = y 3 + C y Therefore, the required solution is x = y 3 + C y .

If y ( t ) is a solution of ( 1 + t ) d y d t - t y = 1 and y ( 0 ) = - 1 , then show that y ( 1 ) = - 1 2 .

Given, ( 1 + t ) d y d t - t y = 1 Dividing both sides by ( 1 + t ) ⇒ d y d t - t 1 + t y = 1 1 + t Which is a linear differential equation. On Comparing it with d y d t + P y = Q , we get P = - t 1 + t , Q = 1 1 + t Now, IF = e - ∫ t 1 + t d t = e - ∫ 1 - 1 1 + t d t = e - [ t - log ( 1 + t ) ] = e - t e log ( 1 + t ) = e - t ( 1 + t ) The general solution is given by y · IF = ∫ Q · IF d x + C y ( t ) · ( 1 + t ) e t = ∫ ( 1 + t ) e - t ( 1 + t ) d t + C = - e - t + C ⇒ y ( t ) = - e - t e t ( 1 + t ) + k , where k = C e t 1 + t ⇒ y ( t ) = - 1 1 + t + k .....(i) Given that y ( 0 ) = - 1 i.e when t = 0 then y = - 1 . Substituting it in Eq.(i), we get - 1 = - 1 + k ⇒ k = 0 Eq.(i) becomes, y ( t ) = - 1 1 + t Now, y ( 1 ) = - 1 2 Hence proved.