Solve (dy/dx) + y cot x = 2 cos x

1.	Solve (dy/dx) + y cot x = 2 cos x
Answer» ∵ (dy/dx) + y cot x = 2 cos x Here (dy/dx) + Py = ϕ is integral equation of linear ∴ P = cot x and ϕ = 2 cos x I.F = e^∫Pdx= e^{∫cot x dx} = e^{log(sin x)} = sin x Now solution of differential equation is y x I.F. = ∫ϕ x I.F dx + c y sin x = ∫2 cos x . sin x + c y sin x = ∫sin 2x + c y sin x = - (cos 2x/2) + c

Answer»

∵ (dy/dx) + y cot x = 2 cos x

Here (dy/dx) + Py = ϕ is integral equation of linear

∴ P = cot x and ϕ = 2 cos x

I.F = e^∫Pdx= e^{∫cot x dx}

= e^{log(sin x)} = sin x

Now solution of differential equation is

y x I.F. = ∫ϕ x I.F dx + c

y sin x = ∫2 cos x . sin x + c

y sin x = ∫sin 2x + c

y sin x = - (cos 2x/2) + c

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Solve (dy/dx) + y cot x = 2 cos x

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