CBSE Class 12 · 2009 · Differential EquationsLong Answer

(a) Solve the following differential equation:

\frac{d y}{d x} + y = cos x - sin x

(b) Find the particular solution, satisfying the given condition, for the following differential equation:

\frac{d y}{d x} - \frac{y}{x} + cosec (\frac{y}{x}) = 0; y = 0 when x = 1

Answer: (a) $y = cos x + C e^{- x}$ ; (b) $cos \frac{y}{x} = lo g ∣ x ∣ + 1$

Solution

(a) Linear with

P = 1

I.F. = e^{x}

y e^{x} = \int e^{x} (cos x - sin x) d x = e^{x} cos x + C

since

\frac{d}{d x} (e^{x} cos x) = e^{x} (cos x - sin x)

. Hence

y = cos x + C e^{- x}

(b)

\frac{d y}{d x} = \frac{y}{x} - cosec \frac{y}{x}

is homogeneous. Put

y = v x

v + x \frac{d v}{d x} = v - cosec v \Rightarrow sin v d v = - \frac{d x}{x}

- cos v = - lo g ∣ x ∣ + C_{1} \Rightarrow cos \frac{y}{x} = lo g ∣ x ∣ + C

Apply

y = 0, x = 1

cos 0 = 0 + C \Rightarrow C = 1

cos \frac{y}{x} = lo g ∣ x ∣ + 1

CBSE Class 12 Board · 2009 · Subjective