Differential EquationMedium
Question
If
ty(t)dt = x2 + y(x) then y as a function of x is-
ty(t)dt = x2 + y(x) then y as a function of x is-Options
A.y = 2 - (2 + a2)e
B.y = 1 - (2 + a2)e
C.y = 2 - (1 + a2)e
D.none
Solution
y(t)dt = x + y(x)⇒ xy = 2x +
⇒ x(y - 2) = 
⇒ ∫xdx =
ln|y - 2| + lnc⇒ ex2/2 = c(y - 2)
at x = a Y = - a2
∴ ex2/2 = c(-a2 - 2) ⇒ c = -
∴
(y - 2)⇒ - y + 2 = (a2 + 2)e
⇒ y = 2 - (2 + a2)e
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