Thread: General Solution to Differential Equation

1. General Solution to Differential Equation

Hi, I would be really grateful to anyone who may help with this...

I need to find the general solution for the differential equation

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

(giving the solution in implicent form)

I have differentiate the function f(x) = e^cos x + sin x using the Composite Rule into (cos x - sin x) e ^ cos x + sin x (Is this correct?)

I also need to find the particular solution of the differential equation for which y = 1 when x = 0 and then give this particular solution in explicit form..

Many thanks for any help!

2. Originally Posted by looking0glass
Hi, I would be really grateful to anyone who may help with this...

I need to find the general solution for the differential equation

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

(giving the solution in implicent form)

I have differentiate the function f(x) = e^cos x + sin x using the Composite Rule into (cos x - sin x) e ^ cos x + sin x (Is this correct?)

I also need to find the particular solution of the differential equation for which y = 1 when x = 0 and then give this particular solution in explicit form..

Many thanks for any help!

$
y' = \left(\cos x - \sin x\right)e^{\sin x + \cos x - y}\;?
$

3. Originally Posted by looking0glass
Hi, I would be really grateful to anyone who may help with this...

I need to find the general solution for the differential equation

dy/dx = (cos x - sin x)e^cos x + sin x - y
If this is $\frac{dy}{dx}= (cos x- sin x)e^{cos x+ sin x- y}$, as danny arigo suggests, then it is separable. $\frac{dy}{dx}= (cos x- sin(x))e^{cos x+ sin x}e^{-y}$ so $e^y dy= (cos x- sin x)e^{cos x + sin x} dx$. Because, as you say, the derivative of $e^{cos x+ sin x}$ is $(cos x- sin x)e^{cos x+ sin x}$, that's easy to integrate.

(giving the solution in implicent form)

I have differentiate the function f(x) = e^cos x + sin x using the Composite Rule into (cos x - sin x) e ^ cos x + sin x (Is this correct?)

I also need to find the particular solution of the differential equation for which y = 1 when x = 0 and then give this particular solution in explicit form..

Many thanks for any help!
Integrating the equation in "separated" form above gives a "constant of integration". Use that fact that f(0)= 1 to determine that constant.

4. Thanks for your help but I'm still not sure how to go about doing this...

5. Originally Posted by looking0glass
Thanks for your help but I'm still not sure how to go about doing this...
As Hallsofivy said, the equation separates so

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

so

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

Let $u = cos x + sin x$ so $du = (cos x- sin x)dx$ so

$
\int e^y dy= \int e^u du\;\;\; \Rightarrow\;\;\; e^y = e^u + c$

or

$
e^y = e^{cos x + sin x} + c$

Now use the fact that $y(0) = 1$ to find $c$. Then take the $\ln$ of both sides to find $y$.

Hope that helps!