Question

# The solution of differential equation cos2xdydx−(tan2x)y=cos4x,|x|<π4, where y(π6)=3√38

A
y=tan2xcos2x
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B
y=cot2xcos2x
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C
2y=tan2xcos2x
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D
2y=cot2xcos2x
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Solution

## The correct option is A y=tan2xcos2xGiven that:- cos2xdydx−(tan2x)y=cos4x,|x|<π4, where y(π6)=3√38To find:- Solution of the given equation.Solution:- ∵cos2x(dydx)−(tan2x)y=cos4x⇒cos2x(dydx)−tan2xy=cos4x⇒dydx−tan2xcos2x.y=cos4xNow, we will compat the eqn i todydx+Py=B∴P=f(x)B=f(x)So, Integration factor (I.F) =e∫PdxHere we will find the value of −tan2xcos2x−∫tan2xcos2xdx=−∫2sin2xcos2x(2cos2x)dx=−∫2sin2xdxcos2x(1+cos2x)Let, t=cos2x⇒dt=−2sinxdx⇒=∫dtt(1+t)=∫1t−11+tdt=ln∣∣∣t1+t∣∣∣=ln∣∣∣cos2x1+cos2x∣∣∣Now, putting ln∣∣∣cos2x1+cos2x∣∣∣ in I.F∴ePdx=eln∣∣∣cos2x1+cos2x∣∣∣=cos2x1+cos2x=cos2x2cos2xSince, multiplying eqn i by I.F both side, we get.cos2x2cos2x.dydx−sin2xcos4xy=cos2x2cos2x2cos2x.dy−(sin2xcos4x)dx.y=cos2x2dx⇒∫d(ycos2x2cos2x)=∫cos2x2dx⇒y.cos2x2cos2x=sin2x4+C...........(II)⇒3√38(1/22.34)=√38+c⇒x=π6y=3√38⇒√38=√38+c∴c=0putting the value of c in eq ii.∴y.cos2x2cos2x=sin2x4+0⇒y=sin2x.2cos2xcos2x.4∴y=tan2x.cos2x2∴2y=tan2x.cos2xhence, the required solution is2y=tan2x.cos2x

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