d/(dθ)(r(θ))+r(θ)tan(θ)=9tan(θ)

解答

\frac{d r}{d θ} + r tan (θ) = 9 tan (θ)

r = \frac{9 tan ( θ )}{1 + tan ( θ ) θ}; 1 + tan (x) x \neq = 0

求解步骤

\frac{d}{d θ} (r θ) + r θ tan (θ) = 9 tan (θ)

\frac{d}{d θ} (r θ) = r

r + r θ tan (θ) = 9 tan (θ)

r + r θ tan (θ) = 9 tan (θ) : r = \frac{9 tan ( θ )}{1 + tan ( θ ) θ}; 1 + tan (x) x \neq = 0

r = \frac{9 tan ( θ )}{1 + tan ( θ ) θ}; 1 + tan (x) x \neq = 0

x^{2} y^{^{'}} = 3 y^{2}

4 y^{^{'}} - e^{3 x} y = 0

y^{^{'}} = 5 x y

u^{^{'}} x + H (x) = \frac{1 - H ( x )}{1 + H ( x )}

x^{2} (x + 1) (2 x^{2} - x - 5) d x = \frac{1}{( y + 1 )} d y