Question

dy/dx= y^4 and y(2)= -1. Y(-1)=

Answer 1

It looks like you're asked to find the value of y(-1) given its implicit derivative,

$\dfrac{dy}{dx} = y^4$

and with initial condition y(2) = -1.

The differential equation is separable:

$\dfrac{dy}{y^4} = dx$

Integrate both sides:

$\displaystyle \int \frac{dy}{y^4} = \int dx$

$-\dfrac1{3y^3} = x + C$

Solve for y :

$\dfrac1{3y^3} = -x + C$

$3y^3 = \dfrac1{-x+C} = -\dfrac1{x + C}$

$y^3 = -\dfrac1{3x+C}$

$y = -\dfrac1{\sqrt[3]{3x+C}}$

Use the initial condition to solve for C :

$y(2) = -1 \implies -1 = -\dfrac1{\sqrt[3]{3\times2+C}} \implies C = -5$

Then the particular solution to the differential equation is

$y(x) = -\dfrac1{\sqrt[3]{3x-5}}$

and so

$y(-1) = -\dfrac1{\sqrt[3]{3\times(-1)-5}} = \boxed{\dfrac12}$