# Thread: solve the given differential by variation of parameters..

1. ## solve the given differential by variation of parameters..

ok so the problem is

xy'' - 4y' = x^4

im getting a different answer from the book and was wondering if someone could check my work.

Yc= c1 + c2x^5

W = 5x^4

W1 = -x^9

W2 = x^4

it seems that the book is using f(x) = x^3???

i thought the f(x) function would be x^4?

2. If you let $\displaystyle \displaystyle Y = y'$ then the DE becomes

$\displaystyle \displaystyle xY' - 4Y = x^4$

$\displaystyle \displaystyle Y' - \frac{4}{x}Y = x^3$.

The integrating factor is $\displaystyle \displaystyle e^{\int{-\frac{4}{x}\,dx}} = e^{-4\ln{x}} = e^{\ln{\left(x^{-4}\right)}} = x^{-4}$. Multiplying both sides of the DE by the integrating factor gives

$\displaystyle \displaystyle x^{-4}Y' - 4x^{-5}Y = x^{-1}$

$\displaystyle \displaystyle \left(x^{-4}Y\right)' = x^{-1}$

$\displaystyle \displaystyle x^{-4}Y = \int{x^{-1}\,dx}$

$\displaystyle \displaystyle x^{-4}Y = \ln{|x|} + C_1$

$\displaystyle \displaystyle Y = x^4\ln{|x|} + C_1x^4$

$\displaystyle \displaystyle y' = x^4\ln{|x|} + C_1x^4$

$\displaystyle \displaystyle y = \int{x^4\ln{|x|} + C_1x^4\,dx}$

$\displaystyle \displaystyle y = \frac{1}{5}x^5\ln{|x|} - \frac{1}{25}x^5 + \frac{C_1}{5}x^5 + C_2$.

3. thanks for that, i got everything except the C1 term.

my answer was y= (1/5)x^5ln|x|-(1/25)x^5+x^5C1+C2 <-- ill take another look at the problem.

4. Originally Posted by slapmaxwell1
thanks for that, i got everything except the C1 term.

my answer was y= (1/5)x^5ln|x|-(1/25)x^5+x^5C1+C2 <-- ill take another look at the problem.
What we've got only differs by a constant, since $\displaystyle \displaystyle \frac{C_1}{5}$ is still a constant...