1. ## seperable equation

ive got a differential equation

y'-y=xy to the power 5,

so have let (ie. .)

Then

Turning this around we get:

So the differential equation becomes:

<-- Multiply both sides by .

and g(x)= exp^(intergral of 4 dx)

so

finally giving y= -x + 1/4 -c/e^{4x}

is this the final answer, and do i need to bring back in the substitution i did earlier?

2. Originally Posted by Student_richard
ive got a differential equation

y'-y=xy to the power 5,

so have let (ie. .)

Then

Turning this around we get:

So the differential equation becomes:

<-- Multiply both sides by .

and g(x)= exp^(intergral of 4 dx)

so
You are good up to here.

And remember, you are currently solving the u(x) equation! Now we have that
$\displaystyle u(x) = \frac{\int e^{4x} \cdot (-4x) dx + C}{e^{4x}}$

$\displaystyle u(x) = \frac{-4 \int xe^{4x}dx + C}{e^{4x}}$

$\displaystyle u(x) = \frac{-4 \left ( \frac{1}{4}xe^{4x}- \frac{1}{16}e^{4x} \right ) + C}{e^{4x}}$

$\displaystyle u(x) = -x + \frac{1}{4} + Ce^{-4x}$

Now, $\displaystyle y = u^{-1/4}$, so

$\displaystyle y(x) = \frac{1}{\sqrt[4]{-x + \frac{1}{4} + Ce^{-4x}}}$

-Dan

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