Particular solutions - DE

[shiri]

Find the particular solution of the differential equation

$$xy' + 4y = -20xcos(x^5)$$

satisfying the initial condition y(pi) = 0.Solution

$$y' + 4(y/x) = -20cos(x^5)$$

p(x) = $$4/x$$
q(x) = $$-20cos(x^5)$$

$$u(x) = int(4/x) = 4lnx$$
$$e^{u(x)} = x^4$$

$$1/e^{u(x)} int(e^{u(x)}*q(x)dx$$
$$1/(x^4) int(x^4*(-20cos(x^5)))dx$$

u substitution
$$u = x^5$$
$$du = 5x^4dx$$

$$1/(x^4) int(x^4*(-20cos(u)))(du/(5x^4))$$
$$-4/(x^4) int(cosu)du$$
$$-4/(x^4) (sin(x^5)+c)$$
$$(-4sin(x^5)/(x^4))-((4c)/(x^4))$$

$$y(pi) = 0 = (-4sin(pi^5)/(pi^4))-((4c)/(pi^4))$$
$$c = -sin(pi^5)$$

Final
$$(-4sin(x^5)/(x^4))+((4sin(pi^5))/(x^4))$$

So this is what I got, but it's a wrong answer. Can anybody tell me what I do wrong in this problem?

 

[tiny-tim]

Hi shiri!

(have a pi: π and an integral: ∫ and try using the X² tag just above the Reply box )

$$xy' + 4y = -20xcos(x^5)$$
…
$$1/(x^4) int(x^4*(-20cos(x^5)))dx$$

Haven't you lost the x in -20xcosx⁵?

 

[shiri]

Hi shiri!

(have a pi: π and an integral: ∫ and try using the X² tag just above the Reply box )


Haven't you lost the x in -20xcosx⁵?

well I divide both sides by x, so...

 

[tiny-tim]

Sorry, you're right …

xy' + 4y = -20xcosx⁵ …

multiply by x³ …

x⁴y' + 4x³y = -20x⁴cosx⁵

So (x⁴y)' = -4(sinx⁵)'

So x⁴y = 4(sinπ⁵ - sinx⁵)

So y = 4(sinπ⁵ - sinx⁵)/x⁴


hmm … the condition y(π) = 0 is a bit strange for cos(x⁵) …

is it possible the question should be (cosx)⁵ ?