dydt=(1−2t)y2dyy2=(1−2t)∗dt−1y=t−t2−Cy=1(C−t+t2)dydt=y2∗sin(t)dyy2=sin(t)∗dt−1y=−cos(t)−Cy=1C+cos(t)\dfrac{dy}{dt}=(1-2t)y^2\newline \dfrac{dy}{y^2}=(1-2t)*dt\newline \dfrac{-1}{y}=t-t^2-C\newline y=\dfrac{1}{(C-t+t^2)}\newline \dfrac{dy}{dt}=y^2*\sin{(t)}\newline \dfrac{dy}{y^2}=\sin{(t)}*dt\newline \dfrac{-1}{y}=-\cos{(t)}-C\newline y=\dfrac{1}{C+\cos{(t)}}dtdy=(1−2t)y2y2dy=(1−2t)∗dty−1=t−t2−Cy=(C−t+t2)1dtdy=y2∗sin(t)y2dy=sin(t)∗dty−1=−cos(t)−Cy=C+cos(t)1
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