Math, asked by prathamsagar32, 10 months ago

y' + y tan x = sin 2x; y (0) = 1.

Answers

Answered by ᎷíssGℓαмσƦσυs
0

dy/dx+ysec²x+tanxdy/dx=2cos2x

dy/dx(1+tanx)=2cos2x-ysec²x

dy/dx=( 2cos2x-ysec²x)/1+tanx

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