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(x^4) dy/dx = (5/3)X^5(y-y^4) sin (x)

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hialex | Student, Undergraduate | eNoter

Posted May 3, 2013 at 12:45 PM via web

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(x^4) dy/dx = (5/3)X^5(y-y^4) sin (x)

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted May 5, 2013 at 5:39 PM (Answer #1)

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Youneed to separate the variables, such that:

`(dy)/(y - y^4) = (x^5*sin x)/x^4 dx`

Reducing duplicate factors yields:

`(dy)/(y - y^4) = (x*sin x)dx`

Integrating both sides yields:

`int (dy)/(y - y^4) = int (x*sin x)dx`

You need to use partial fraction decomposition to evaluate `int (dy)/(y - y^4)` , such that:

`1/(y(1 - y^3)) = a/y + b/(1 - y) + (cy + d)/(1 + y + y^2)`

`1 = a(1 - y^3) + b(y + y^2 + y^3) + (cy + d)(y - y^2)`

`1 = a - ay^3 + by + by^2 + by^3 + cy^2 - cy^3 + dy - dy^2`

`1 = y^3(b - a - c) + y^2(b + c - d) + y(b + d) + a`

Equating the coefficients of like powers yields:

`b - a - c = 0`

`b + c - d = 0`

`b + d = 0`

`a = 1 => {(b - c = 1),(2b + c = 0):} => 3b = 1 => b = 1/3 => d = -1/3`

`c = 1/3 - 1=> c = -2/3`

`1/(y(1 - y^3)) = 1/y + 1/(3(1 - y)) - (2y + 1)/(3(1 + y + y^2))`

Integrating both sides yields:

`int 1/(y(1 - y^3)) dy = int 1/y dy + int 1/(3(1 - y)) dy - int (2y + 1)/(3(1 + y + y^2)) dy`

`int 1/(y(1 - y^3)) dy = ln |y| - (1/3)ln|1 - y| - (1/3) ln(y^2 + y +1) + c`

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

`int 1/(y(1 - y^3)) dy = ln |y| - ln root(3)|(1 - y^3)| + c`

`int 1/(y(1 - y^3)) dy = ln |y/(root(3)(1 - y^3))| + c`

You need to integrate to the right such that:

`int x*sin x dx = -x*cos x + int cos x dx`

`u = x => du = 1`

`dv = sin x => v = -cos x`

`int x*sin x dx = -x*cos x + sin x + c`

`int 1/(y(1 - y^3)) dy = int x*sin x dx => ln |y/(root(3)(1 - y^3))| = -x*cos x + sin x + c`

`y/(root(3)(1 - y^3)) = e^(-x*cos x + sin x + c)`

`y^3/(1 - y^3) = e^(3(-x*cos x + sin x + c))`

Hence, evaluating the intregrals, using the separation of variables, yields `y^3/(1 - y^3) = e^(3(-x*cos x + sin x + c)).`

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