# Thread: Silly De questions to muck up my brain!

1. ## Silly De questions to muck up my brain!

I have two questions, trying to learn to do DE's using substitution....my problems ussally come out looking like horrbile bog monsters of intergation (like a parts of x^9 (e^x)) i could intergrate it but id like to keep my work to under 3 pages .....anyway:

i have a (dy/dx) = cos (x+y)

and

dy/dx - y/x = y^3

also i have one wth the dentotation "any appropriate method" and i trie excat and came out with the above explained bog monster....

it starts ...(x(dy/dx))-4y=x^6(e^x)

Thanks agian all, any help would be appreciated....

2. Originally Posted by neven87
I have two questions, trying to learn to do DE's using substitution....my problems ussally come out looking like horrbile bog monsters of intergation (like a parts of x^9 (e^x)) i could intergrate it but id like to keep my work to under 3 pages .....anyway:

i have a (dy/dx) = cos (x+y)
Put u=x+y, then du/dx=1+dy/dx so the original equation becomes:

du/dx -1 = cos(u),

which is of variables seperable type:

integral 1/(cos(u) +1) du = integral dx

RonL

3. Hello, neven87!

The last one can be solved with an integrating factor.

$\displaystyle x\frac{dy}{dx} - 4y \:=\:x^6e^x$

Divide by $\displaystyle x\!:\;\;\frac{dy}{dx} - \frac{4}{x}y \:=\:x^5e^x$

Then: .$\displaystyle I \;=\;e^{-4\int\frac{dx}{x}} \;=\;e^{-4\ln x} \;=\;e^{\ln(x^{-4})} \;=\;x^{-4}\;=\;\frac{1}{x^4}$

Multiply by $\displaystyle I\!:\;\;\frac{1}{x^4}\frac{dy}{dx} - \frac{4}{x^5} \;=\;xe^x$

And we have: .$\displaystyle \frac{d}{dx}\left(\frac{1}{x^4}y\right) \;=\;xe^x$

Integrate: .$\displaystyle \int \frac{d}{dx}\left(\frac{1}{x^4}y\right) \;=\;\underbrace{\int xe^x\,dx}_{\text{by parts}}$

And we have: .$\displaystyle \frac{1}{x^4}y \;=\;e^x(x - 1) + C$

Therefore: .$\displaystyle y \;=\;e^xx^4(x-1) + Cx^4$