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Question

Solution of differential equation siny.dydx+1xcosy=x4cos2y is

A
xsecy=x6+C
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B
6xsecy=x+C
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C
6xsecy=x6+C
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D
6xsecy=6x6+C
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Solution

The correct option is D 6xsecy=x6+C
siny×dydx+1xcosy=x4cos2y

divide both side by siny

=>dydx+1xcosysiny=x4cos2ysiny

=>dydx+1xcoty=x4cotycosy

Now this in the form of

dydx+py=q(i)

p=1x,q=x4cotycosy

first we solve Integreted factor

=>e1xdx

=>e(logx)

=>x

now put the value if p and d in equation (i) we get,

=>y×x=x4×coty×cosx×xdx

=>xy=x5cotycosxdx


=>xy=x6cotycosy6

=>6xsecy=x6+c

























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