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CBSE Class 12 · 2014 · Differential EquationsLong Answer

Find the particular solution of the differential equation

\frac{d y}{d x} = 1 + x + y + x y

, given that

y = 0

when

x = 1

.

Answer: $y = e^{(x^{2} + 2 x - 3) /2} - 1$

Solution

Factor the right side:

\frac{d y}{d x} = (1 + x) (1 + y)

.

\frac{d y}{1 + y} = (1 + x) d x \Rightarrow lo g ∣1 + y ∣ = x + \frac{x ^{2}}{2} + C

Apply

y = 0, x = 1

:

0 = 1 + \frac{1}{2} + C \Rightarrow C = - \frac{3}{2}

. Thus

lo g ∣1 + y ∣ = \frac{x ^{2} + 2 x - 3}{2}

:

y = e^{(x^{2} + 2 x - 3) /2} - 1

CBSE Class 12 Board · 2014 · Subjective

Practise more: all CBSE board PYQs on Differential Equations →

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