Thread: Solve the Differential Equation

1. Solve the Differential Equation

sec(x) dy/dx = e^(y + sin(x))

So far I have :

ln (secx) dy/dx = y + sin x

ln (secx) dy = y + sinx dx

Not sure what to do now... I don't think I can just subtract the y?

2. you should be able to solve this using implicit differentiation.

so you have $\displaystyle \frac{dy}{dx} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}} = -\frac{F_x}{F_y}$

so then

$\displaystyle \sec(x) dy/dx = e^{y + \sin(x)}$

$\displaystyle = \frac{1}{\cos(x)} \frac{dy}{dx} = e^{y + \sin(x)}$

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

$\displaystyle \frac{\partial F}{\partial y} = e^{y + \sin(x)} \times \cos(x)$

$\displaystyle \frac{\partial F}{\partial x} = \cos(x) \times e^{y + \sin(x)} \times \cos(x) -sin(x) \times e^{y + \sin(x)} = e^{y + \sin(x)}(\cos^2(x) -\sin(x))$

now just combine both function...

3. Originally Posted by veronicak5678 sec(x) dy/dx = e^(y + sin(x))

So far I have :

ln (secx) dy/dx = y + sin x

ln (secx) dy = y + sinx dx

Not sure what to do now... I don't think I can just subtract the y?
Note that $\displaystyle \sec(x)\frac{\,dy}{\,dx}=e^{y+\sin x}\implies\sec(x)\frac{\,dy}{\,dx}=e^{y}e^{\sin x}\implies e^{-y}\,dy=\frac{e^{\sin x}}{\sec x}\,dx$ $\displaystyle \implies e^{-y}\,dy=e^{\sin x}\cos x \,dx$

Can you take it from here?

--Chris

4. I see. Thanks!

5. Originally Posted by veronicak5678 I see. Thanks!
$\displaystyle e^{-y}$, because in simplifying, I had $\displaystyle \frac{1}{e^y}\,dy=e^{\sin x}\cos x\,dx$, but $\displaystyle \frac{1}{e^y}=e^{-y}$

Does this clarify things?

--Chris

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cosx dy/dx ysinx = sec.secx

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