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The Integral function of differential equation dy/dx - y + tanx = cosx is

Answer»

\(\frac{dy}{dx}\) - y + tanx = cosx

I.F = e ∫-tanx dx

= e-log secx = \(\frac{1}{secx}=cosx\)

It's complete solution is

y x (I.F) = ∫(I.F) x 2 dx

\(\Rightarrow\) y. cosx = ∫cos2x dx

= \(\int \frac{1+cos2x}{2}\,dx\)

= \(\frac{x}{2}+\frac{sin\,2x}{4}+c\)

Hence, complete solution of given differential equation is y = \(\frac{x}{2}\) sec x + \(\frac{sin x}{2}+c\) sec x.



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