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2 years ago

11

Differential cos^2x dy/dx =e^y-tanx

1 answer:

gulaghasi [49]2 years ago

8 0

Answer:

y=t−1+ce

−t

where t=tanx.

Given, cos

2

x

dx

dy

+y=tanx

⇒

dx

dy

+ysec

2

x=tanxsec

2

x ....(1)

Here P=sec

2

x⇒∫PdP=∫sec

2

xdx=tanx

∴I.F.=e

tanx

Multiplying (1) by I.F. we get

e

tanx

dx

dy

+e

tanx

ysec

2

x=e

tanx

tanxsec

2

x

Integrating both sides, we get

ye

tanx

=∫e

tanx

.tanxsec

2

xdx

Put tanx=t⇒sec

2

xdx=dt

∴ye

t

=∫te

t

dt=e

t

(t−1)+c

⇒y=t−1+ce

−t

where t=tanx

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