Find the explicit solution of the following initial value problem. xy^2 dy/dx=y^3-x^3   ,   y(1)=2

Expert Answers

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Start by dividing by y^3 both sides:

`(x/y)*(dy/dx) = 1 - (x/y)^3`

Put `x/y = t =gt x= ty =gt dx = (t'*y + t)dt =gt dt/dx = 1/(t'*y + t)`

`` `t/(t'*y + t) = 1 - t^3 =gt t'*y + t = t/(1 - t^3)`

`y*(dt/dy) = t/(1 - t^3) - t`  => `t*(dt/dy) = -t^4/(1 - t^3)`

Integrate both sides:

`int dy/y =- int (1 - t^3)/t^4`

`` `ln |y| = - t^-3/-3 + ln |t|`

`ln |y| = -1/(3t^3)- ln |t| + c`

`` `ln |y|+ ln |t| = -1/3t^3 + c`

`` `ln |yt| = -1/3t^3 + c`

`` `ln |x| = -1/3(x/y)^3 + c`  =>`x = 1/e^(3(x/y)^3)`

ANSWER: `x = 1/e^(3(x/y)^3)`

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