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

User Akkishore
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1 Answer

9 votes

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}

User JayNCoke
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