A curve is such that dy/dx=8/(5-2x)^2. Given that the curve passes through 2,7 Find the equation of the curve. I want to know that how do I know i have to solve it through integration or differentiation etc how to solve it?plz tell when to integrate

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Answer 1 · 2017-11-26T06:35:24

$y=4/(5-2x)+3$

As $(dy)/(dx)=8/(5-2x)^2$

$y=intdy=int8/(5-2x)^2dx$

Let $u=5-2x$ then $du=-2dx$

and $int8/(5-2x)^2dx=int8/u^2(-(du)/2)$

= $-4int(du)/u^2$

= $-4(u^(-1)/-1)+c$

= $4/u+c$

i.e. $y=4/(5-2x)+c$

As curve passes through $(2,7)$ , for $x=2$ we have $y=7$

therefore $7=4/(5-4)+c$ or $c=3$

Hence $y=4/(5-2x)+3$

graph{4/(5-2x)+3 [-8.87, 11.13, -1.48, 8.52]}