Question

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

Answer 1

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

Explanation:

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]}