What is the solution for (dr)/(dθ)=-pisin(piθ),r(2)=3 ?

The solution for (dr)/(dθ)=-pisin(piθ),r(2)=3 is r=cos(piθ)+2

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(dr)/(dθ)=-pisin(piθ),r(2)=3

\frac{d r}{d θ} = - π sin (π θ), r (2) = 3

r = cos (π θ) + 2

Solution steps

\frac{d r}{d θ} = - π sin (π θ)

Substitute

\frac{d r}{d θ}

with

r^{'} (θ)

r^{^{'}} (θ) = - π sin (π θ)

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

r = \int - π sin (π θ) d θ

\int - π sin (π θ) d θ = cos (π θ) + c_{1}

r = cos (π θ) + c_{1}

Apply initial conditions:

r = cos (π θ) + 2

r = cos (π θ) + 2

What is the solution for (dr)/(dθ)=-pisin(piθ),r(2)=3 ?
The solution for (dr)/(dθ)=-pisin(piθ),r(2)=3 is r=cos(piθ)+2