dy/dt= 0.5(y^2)-2y where y=0 for x=0
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Originally Posted by fastman390 dy/dt= 0.5(y^2)-2y where y=0 for x=0 $\displaystyle \int\frac{dy}{y^2- 4y}= \int 0.5 dx$ $\displaystyle y^2- 4y= y(y-4)$ so you can do the left side by "partial fractions" as your title says.
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