What is the solution for (dr)/(dθ)=pisin(piθ),r(1)=3 ?

The solution for (dr)/(dθ)=pisin(piθ),r(1)=3 is r(θ)=-cos(piθ)+2

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(dr)/(dθ)=pisin(piθ),r(1)=3

\frac{d r}{d θ} = π sin (π θ), r (1) = 3

r (θ) = - 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 (θ) = \int π sin (π θ) d θ

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

r (θ) = - cos (π θ) + c_{1}

Apply initial conditions:

r (θ) = - cos (π θ) + 2

r (θ) = - cos (π θ) + 2

(- \frac{4}{3}) \cdot e^{- 2 t} + (- \frac{2}{3}) \cdot e^{t}

cos (t + 1)

\int \frac{x ^{2} + x}{x ^{2} + x + 1} d x

\int \frac{1}{7 - 4 x} d x

y = e^{- 11 x}

What is the solution for (dr)/(dθ)=pisin(piθ),r(1)=3 ?
The solution for (dr)/(dθ)=pisin(piθ),r(1)=3 is r(θ)=-cos(piθ)+2