If dydx+3cos2xy =1cos2x, x∈(−π3, π3) and y(π4) = 43 ,then y(−π4) equal
13
13+e6
13+e3
−43
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dydx+(3sec2x)y=sec2H
This is linear differential equation
Hence y.e∫3tanx.sec2xdx
⇒ye3tanx=e3tanx3+c
⇒y=Ce−3tanx+13
Giveny y(π4)=43⇒43=Ce−3+13
⇒ C = e3
Hence y(π4)=e3.e3+13=e6+13