(dy)/(dθ)=sin^4(piθ)

解答

\frac{d y}{d θ} = sin^{4} (π θ)

y = \frac{1}{π} (- \frac{1}{4} sin^{3} (π θ) cos (π θ) + \frac{3}{8} (π θ - \frac{1}{2} sin (2 π θ))) + c_{1}

求解步骤

\frac{d y}{d θ} = sin^{4} (π θ)

替代

\frac{d y}{d θ}

与

y^{'} (θ)

y^{^{'}} (θ) = sin^{4} (π θ)

若

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

则

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

y = \int sin^{4} (π θ) d θ

\int sin^{4} (π θ) d θ = \frac{1}{π} (- \frac{1}{4} sin^{3} (π θ) cos (π θ) + \frac{3}{8} (π θ - \frac{1}{2} sin (2 π θ))) + c_{1}

y = \frac{1}{π} (- \frac{1}{4} sin^{3} (π θ) cos (π θ) + \frac{3}{8} (π θ - \frac{1}{2} sin (2 π θ))) + c_{1}

(4 y + y x^{2}) d y - (2 x + x y^{2}) d x = 0

d er i v a t i v e lo g_{2} (x lo g_{5} (x))

\int x^{2} sin (5 x) d x

\int (\frac{3 w ^{2}}{2} - \frac{2}{3 w ^{2}}) d w

\frac{d}{d x} (ln^{4} (x^{3} + 2 x - e^{- 3 x}))