The differential equation of y = c 2 + c x is .......

d 2 y d x 2 + x d y d x + y = 0

The solution of 1 x · d y d x = tan - 1 x is ...

y = x 2 tan - 1 x 2 - 1 2 x - tan - 1 x + c

Form the differential equation by eliminating the arbitrary constants A and B from A x 2 + B y 2 = 1 .

Given, A x 2 + B y 2 = 1 Differentiate both sides w.r.t x , we get 2 A x + 2 B y · d y d x = 0 ⇒ 2 B y · d y d x = - 2 A x ⇒ y x d y d x = - A B Again differentiate both sides w.r.t x , we get y x d 2 y d x 2 + x d y d x - y x 2 · d y d x = 0 ⇒ x y d 2 y d x 2 + x d y d x - y d y d x = 0 [Multiply x 2 on both sides] ⇒ x y d 2 y d x 2 + x d y d x 2 - y d y d x = 0 which is the required differential equation.

Form differential equation representing the family of lines making equal intercepts on the co-ordinate axes.

Equation of line in intercept form is x a + y b = 1 where a = x intercept and b = y intercept Given that a = b ⇒ x a + y a = 1 ⇒ x + y = a Differentiating with respect to x on both sides ⇒ d d x x + y = d d x a ⇒ 1 + d y d x = 0 ⇒ d y d x = - 1 which is the required differential equation.

Form the differential equation representing the family of curves y = a sin ( x + b ) , where a , b are arbitrary constants.

Given that, the equation of family of the curve is y = a sin ( x + b ) → 1 Since the given curve is having two arbitrary constants i.e., a and b Therefore, the curve has to be differentiated twice to form the differential equation. Differentiating both sides, we get: d y d x = a cos x + b Again differentiating both sides, we get: d 2 y d x 2 = - a sin x + b ⇒ d 2 y d x 2 = - a × y a [Using equation 1 ] ⇒ d 2 y d x 2 + y = 0 Hence, the required differential equation is d 2 y d x 2 + y = 0 .

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

Given differential equation is d y d x = x 2 y + y ⇒ d y d x = y 1 + x 2 ⇒ d y y = 1 + x 2 d x Integrating both sides, we get ∫ d y y = ∫ 1 + x 2 d x Using ∫ d x a x + b = 1 a log a x + b + c , ∫ x n d x = x n + 1 n + 1 + c , n ≠ - 1 ⇒ log y = x + x 3 3 + c Hence, the required solution of differential equation is log ⁡ y = x + x 3 3 + c .

The rate of growth of bacteria is proportional to the number present. If initially, there were 1000 bacteria and the number doubles in 1 hour, find the number of bacteria after 2 1 2 hours.

Let ' x ' be the number of bacteria present at time ' t ' ∴ d x d t ∝ x ⇒ d x d t = k x , where k is the constant of proportionality. ⇒ d x x = k d t Integrating both sides, we get log x = k t + c . . . . i When t = 0 , we have x = 1000 ⇒ log 1000 = k 0 + c ⇒ c = log ( 1000 ) ∴ log x = k t + log ( 1000 ) … (ii)[From (i) ] Now, when t = 1 , x = 2000 ⇒ log 2000 = k + c [Using i ] ⇒ log 2000 = k + log 1000 ⇒ log ( 2000 ) - log ( 1000 ) = k ⇒ k = log 2000 1000 = log 2 . . . . . i i i [Since log M - log N = log M N ∀ M , N > 0 ] And, when t = 5 2 , from i i , and i i i , we have log x = 5 2 log 2 + log ( 1000 ) = log 2 5 2 + log ( 1000 ) [Since log M p = p log M , ∀ > 0 , p ∈ R ] = log ( 4 2 ) + log ( 1000 ) = log ( 4000 2 ) = log ( 4000 × 1 . 414 ) ∴ log x = log ( 5656 ) Hence, x = 5656 Thus, there will be 5656 bacteria after 5 2 hours.

Find the order and degree of the differential equation. d 2 y d x 2 - 7 d y d x 3 + 6 y = 0

Given D.E is d 2 y d x 2 - 7 d y d x 3 + 6 y = 0 We know that the degree of a differential equation is the power of its highest derivative, after the equation has been made rational and integral in all of its derivatives and the order of a differential equation is the order of the highest order derivative involved in the differential equation. So, order = 2 and degree = 1