\[ a\sin(t)\frac{dx}{dt} -x = 0 \Leftrightarrow a\sin(t)\frac{dx}{dt} = x \\ \Leftrightarrow a\sin(t)dx = xdt \Leftrightarrow \frac{dx}{x}= \frac{dt}{a\sin(t)} = \frac{1}{a}\csc(t) dt \\ \Leftrightarrow \int \frac{dx}{x} = \int \frac{1}{a}\csc(t) dt \Leftrightarrow ln|x| + C = \frac{1}{a} \int \csc(t) \cdot \frac{csc(t)-cot(t)}{csc(t)-cot(t)} dt \\ \Leftrightarrow ln|x| + C = \frac{1}{a} \int \frac{(csc^2(t)-cot(t)csc(t))dt}{csc(t)-cot(t)} \]
Sei nun \( u=csc(t)-cot(t) \Rightarrow \frac{du}{dt}=(-cot(t)csc(t)-(-csc^2(t))) \Rightarrow du=(csc^2(t)-cot(t)csc(t))dt \)
\[ \Rightarrow ln|x| + C = \frac{1}{a} \int \frac{(csc^2(t)-cot(t)csc(t))dt}{csc(t)-cot(t)} \\ \Leftrightarrow ln|x| + C = \frac{1}{a} \int \frac{du}{u} \\ \Leftrightarrow ln|x| + C = \frac{1}{a}ln|u| = \frac{1}{a}ln|csc(t)-cot(t)| \\ \Leftrightarrow e^{ln|x| + C} = e^{\frac{1}{a}ln|csc(t)-cot(t)|} \\ \Leftrightarrow Kx = (csc(t)-cot(t))^{\frac{1}{a}} \\ \Leftrightarrow x(t) = k \sqrt[a]{csc(t)-cot(t)} \]
Punkte: 90