in implicit form find the general solution of the differential equation!!!

(Thinking)

hi,

d/dx= (1+e^{4x})/(4x+e^{4x}) = 1/(8(4x+e^{4x})^{2}) + c

find in implicit for the general solution of the differential equation::

dy/dx= - (4e^{-2y})(1+e^{4x})/(4x+e^{4x})^{3 (x>0) }

help, i realize that the (4e^{-2y}) needs to be on the left hand side, but i keep getting the blooming thing wrong.......and obviously the differentiation above it is heavily involved...... i think i need to actually have a face to face tutorial, but i live 2 hours travel from the tutorial place which would be ok if it wasn't in central london arghh!!!

any help as in super step by step would be appreciated...

many thanks

Re: in implicit form find the general solution of the differential equation!!!

I would ask that you proofread what you wrote, because the very first line makes no sense. You have d/dx, but you don't say what you are differentiating. So you say this operator is equal to the first expression which is equal to the second. NONE of those things are equal!

Re: in implicit form find the general solution of the differential equation!!!

Hi sorry should read dy/dx not d/dx ...

Re: in implicit form find the general solution of the differential equation!!!

Your first line still doesn't make any sense though, these things are NOT equal...

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Re: in implicit form find the general solution of the differential equation!!!

hi i have attached the question sheet, equation 2.3 is: one of two integration formulas. thank you,

Re: in implicit form find the general solution of the differential equation!!!

$\displaystyle \displaystyle \begin{align*} \int{\frac{1 + e^{4x}}{ \left( 4x + e^{4x} \right) ^3 } \,dx} &= \frac{1}{4} \int{\frac{4 + 4e^{4x}}{\left( 4x + e^{4x} \right) ^3}\,dx} \end{align*}$

Now let $\displaystyle \displaystyle u = 4x + e^{4x} \implies du = \left( 4 + 4e^{4x} \right) dx $ and the integral becomes

$\displaystyle \displaystyle \begin{align*} \frac{1}{4}\int{\frac{4 + 4e^{4x}}{ \left( 4x + e^{4x} \right) ^3} \,dx} &= \frac{1}{4}\int{\frac{1}{u^3}\,du} \\ &= \frac{1}{4} \int{ u^{-3} \,du} \\ &= \frac{1}{4} \left( -\frac{1}{2} u^{-2} \right) + C \\ &= -\frac{1}{8 \left( 4x + e^{4x} \right) ^2 } + C \end{align*}$

Now to solve the DE...

$\displaystyle \displaystyle \begin{align*} \frac{dy}{dx} &= -\frac{4e^{-2y} \left( 1 + e^{4x} \right) }{ \left( 4x + e^{4x} \right) ^3} \\ -\frac{1}{4}e^{2y}\,\frac{dy}{dx} &= \frac{1 + e^{4x}}{ \left( 4x + e^{4x} \right) ^3} \\ \int{-\frac{1}{4}e^{2y}\,\frac{dy}{dx} \, dx} &= \int{ \frac{1 + e^{4x}}{ \left( 4x + e^{4x} \right) ^3} \,dx} \\ -\frac{1}{4} \int{e^{-2y}\,dy} &= \int{ \frac{1 + e^{4x}}{ \left( 4x + e^{4x} \right) ^3} \,dx } \end{align*}$

I trust you can go from here...