Engineering
Mathematics
Linear Differential Equation
Question

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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Solution

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