a gradient of a curve at point (x,y) is given by sin x-y the point (pai, 1) lies on the curve. determine the equation of the curve in the form of y = f(x)

dy/dx=sin(x)-y, dy/dx+y=sin(x).

Multiply through by eˣ:

eˣdy/dx+yeˣ=eˣsin(x)

d(yeˣ)/dx=eˣsin(x)

Integrate:

yeˣ=∫eˣsinxdx+k where k is a constant.

Let u=eˣ, dv=sinxdx then du=eˣ, v=-cosx. Let J=∫eˣsinxdx=-eˣcosx+∫eˣcosxdx.

Let dv=cosxdx, then v=sinx, and ∫eˣcosxdx=eˣsinx-∫eˣsinxdx=eˣsinx-J.

So J=-eˣcosx+eˣsinx-J, 2J=eˣ(sinx-cosx), J=½eˣ(sinx-cosx)=yeˣ-k.

But when x=π, y=1, so k=e^π-½e^π(-(-1))=½e^π.

yeˣ=½eˣ(sinx-cosx)+½e^π

y=f(x)=½(sinx-cosx)+½e^(π-x).

CHECK:

dy/dx=½(cosx+sinx)-½e^(π-x)

y=½(sinx-cosx)+½e^(π-x)

Add these equations together:

dy/dx+y=sinx so this confirms the solution.

And when x=π, y=½+½=1.

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