Q.1
Solution Explain:
Q.2
Solution Explain:
The given differential equation can be written as f(x)dy + f '(x)ydx = dx
i.e. d(f(x).y) = d(x)
Integrating we get y.f(x) = x + c
or f(x)
Q.3
Solution Explain:
Q.4
Solution Explain:
Q.5
Solution Explain:
Q.6
Solution Explain:
Let 
, we have
p2 + y = xp
⇒ 2 pp' + p = p + xp'
⇒ 2p = x or p' = 0
which givens x2 = 4y or y = cx + (2c +1) or y -1 = c(x + 2)
Q.7
Solution Explain:
Q.8
Solution Explain:
Multiplying the first DE by gh, the second by fh and the third by fg, and adding the equations gives
(fgh)′ = 6(fgh)2 + 6
Let f(x) g(x) h(x) = k(x)
We have k′ (x) = 6(k(x))2 + 6
Integrating and using k (0) = 1 gives
Integrating and using f(0) = 1 gives option A and C
Q.9
Solution Explain:
Q.10
Solution Explain:
If y = f(x) is the curve, then
Y – y = f '(x) (X – x) is the equation of the tangent of (x, y). Take f '(x) = dy/dx
Putting X = 0, the initial ordinate of the tangent is Y = y – x f '(x) = subnormal at this point
⇒ y dy/dx ⇒ y – x dy/dx = y dy/dx
which is homogenous of first degree and linear in x.
Q.11
Solution Explain:
If the curve traced by P is y = f(x) then dy/dx = slope of line joining P and Q as P is always moving towards Q
Let Q ≡ (0, β) and P(x, y), (0 – x)2 + (β – y)2 = k2
Solving the differential equation with initial condition f(K) = 0 gives
