# Thread: General solution of eqn in implicit form

1. ## General solution of eqn in implicit form

Hi

I have this eqn and dont know where to begin

$\displaystyle \frac{dy}{dx}=(cos \ x - sin \ x)e^{cos \ x + sin \ x - y}$

I'm asked to find the general solution in implicit form?

Thanks

2. Hello, bobred!

Believe it or not, we can separate the variables . . .

$\displaystyle \frac{dy}{dx}=(\cos x - \sin x)e^{(\cos x + \sin x - y)}$

Find the general solution in implicit form.

We have: .$\displaystyle \frac{dy}{dx} \;=\;(\cos x-\sin x)e^{(\cos x + \sin x)}\cdot e^{-y}$

Separate variables: .$\displaystyle e^y\,dy \;=\;(\cos x - \sin x)e^{(\cos x + \sin x)}$

Integrate: .$\displaystyle \int e^y\,dy \;=\;\int(\cos x - \sin x)e^{(\cos x + \sin x)}\,dx$

On the right, let: $\displaystyle u \:=\:\cos x + \sin x \quad\Rightarrow\quad du \:=\:(\text{-}\sin x + \cos x)dx \quad\Rightarrow\quad (\cos x-\sin x)dx \:=\:du$

The right side becomes: .$\displaystyle \int e^u\,du \:=\:e^u + C \quad\Rightarrow\quad e^{(\cos x + \sin x)} + C$

The equation becomes: .$\displaystyle e^y \;=\;e^{(\cos x + \sin x)} + C$ . . . . implicit form

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We have: .$\displaystyle e^y \;=\;e^{(\cos x + \sin x)} + C$ . . . . let $\displaystyle C \,=\,e^k$

Then we have: .$\displaystyle e^y \;=\;e^{(\cos x+\sin x)} + e^k \quad\Rightarrow\quad e^y \;=\;e^{(\cos x + \sin x + k)}$

Therefore: .$\displaystyle y \;=\;\cos x + \sin x + k$ . . . . explicit form

3. As soon as you said separate the variables and I saw

$\displaystyle e^{(cos \ x + sin \ x)}\cdot e^{-y}$

I got it.

Thanks