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CBSE Class 12 · 2018 · Differential EquationsLong Answer

Find the particular solution of the differential equation

e^{x} tan y d x + (2 - e^{x}) sec^{2} y d y = 0

, given that

y = \frac{π}{4}

when

x = 0

.

Answer: $tan y = 2 - e^{x}$

Solution

Separate the variables:

\frac{sec ^{2} y}{tan y} d y = \frac{- e ^{x}}{2 - e ^{x}} d x = \frac{e ^{x}}{e ^{x} - 2} d x

Each side is a

\frac{f ^{'}}{f}

form:

lo g ∣ tan y ∣ = lo g ∣ e^{x} - 2∣ + lo g C \Rightarrow tan y = C (e^{x} - 2)

Apply

y = \frac{π}{4}, x = 0

:

1 = C (1 - 2) \Rightarrow C = - 1

. Hence

tan y = 2 - e^{x}

CBSE Class 12 Board · 2018 · Subjective

Practise more: all CBSE board PYQs on Differential Equations →

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