1 Answer

JayNCoke · Answer 1 · 2023-11-27T04:41:55+0000

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

(dy)/(dx) = y^4

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

The differential equation is separable:

(dy)/(y^4) = dx

Integrate both sides:

$\displaystyle \int (dy)/(y^4) = \int dx$

$-\frac1{3y^3} = x + C$

Solve for y :

$\frac1{3y^3} = -x + C$

$3y^3 = \frac1{-x+C} = -\frac1{x + C}$

$y^3 = -\frac1{3x+C}$

$y = -\frac1{\sqrt[3]{3x+C}}$

Use the initial condition to solve for C :

$y(2) = -1 \implies -1 = -\frac1{\sqrt[3]{3*2+C}} \implies C = -5$

Then the particular solution to the differential equation is

$y(x) = -\frac1{\sqrt[3]{3x-5}}$

and so

$y(-1) = -\frac1{\sqrt[3]{3*(-1)-5}} = \boxed{\frac12}$

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