Question

Solve the following differential equation :

`"dy"/"dx" + "y" = cos"x" - sin"x"`

Answer 1

Given:
`"dy"/"dx" + "y" = cos"x" - sin"x"` ............(1)

This differential equation is a linear differential equation of the form `"dy"/"dx"+"PQ"="Q"`

P = 1, Q  = cos x  - sin x

I.F. = e^∫Pdx = e^∫1dx = e^x

Now multiply (1) with the I.F. we get

`"e"^"x"("dy"/"dx"+"y") = "e"^"x"(cos"x"-sin"x")`

Integrating both sides with respect to x.
yex=∫excosx-sinxdx+C⇒yex=∫excosxdx-∫exsinxdx+C⇒yex=excosx-∫-sinxexdx-∫exsinxdx+C⇒yex=excosx+∫exsinxdx-∫exsinxdx+C⇒yex=excosx+C">

ye^x = ∫e^x (cos x -sin s)dx + C

⇒ ye^x = ∫e^x cos xdx - ∫e^xsin xdx + C

⇒ ye^x = e^x cosx - ∫(- sinx)e^xdx - ∫e^xsin xdx + C

⇒ ye^x = e^x cosx + ∫e^xsin xdx - ∫e^xsin xdx + C

⇒ ye^x = e^x cosx + C

Thus, ye^x = e^x cosx + C is the required solution of the given differential equation.