Solve the Differential Equation

**veronicak5678** · October 20th 2008, 08:23 PM

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?

**lllll** · October 20th 2008, 08:44 PM

you should be able to solve this using implicit differentiation.

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

so then

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

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

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

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

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

now just combine both function...

**Chris L T521** · October 20th 2008, 08:45 PM

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

Can you take it from here?

--Chris

**veronicak5678** · October 20th 2008, 08:58 PM

I see. Thanks!

**Chris L T521** · October 20th 2008, 09:01 PM

Originally Posted by veronicak5678

I see. Thanks!

$e^{-y}$ , because in simplifying, I had $\frac{1}{e^y}\,dy=e^{\sin x}\cos x\,dx$ , but $\frac{1}{e^y}=e^{-y}$

Does this clarify things?

--Chris

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