`dy/dx = x^2sqrt(x-3)` Solve the differential equation.

Given to solve,

`dy/dx = x^2(sqrt(x-3))`

`dy = x^2(sqrt(x-3)) dx`

on integrating on both sides we get

`y= int x^2(sqrt(x-3)) dx`

so ,

let `u = x-3`

=> `x=u+3 and du =dx`

so,

y= `int x^2(sqrt(x-3)) dx`

=` int (u+3)^2 (u)^(1/2)du`

= `int (u^2+6u+9) *(u^(1/2)) du`

= `int [u^(5/2) + 6u^(3/2) +9u^(1/2)] du`

= `u^((5/2) +1)/((5/2)+1) + 6 u^((3/2) +1)/((3/2)+1)+9u^((1/2) +1)/((1/2)+1) +c`

= `u^((7/2))/((7/2)) + 6 u^((5/2) )/((5/2))+9u^((3/2) )/((3/2)) +c`

=`(x-3)^((7/2))/((7/2)) + 6 (x-3)^((5/2) )/((5/2))+9(x-3)^((3/2) )/((3/2)) +c`

=`2(x-3)^((7/2))/(7) + 12 (x-3)^((5/2) )/(5)+6(x-3)^((3/2) ) +c`

Comments

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